Instructions
Write the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.
When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";
When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;
In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;
If a complex function is given, then it is necessary to multiply the derivative of internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).
Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:
y=x^4, y"=4*x^(4-1)=4*x^3;
y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).
2) Calculate the value of the function at a given point y"(1)=8*e^0=8
Video on the topic
Learn the table of elementary derivatives. This will significantly save time.
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So, what is the difference between an irrational equation and a rational one? If the unknown variable is under the sign square root, then the equation is considered irrational.
Instructions
The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore this equation has no roots.
So, an irrational equation is solved using the method of squaring both its sides. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.
Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, to the right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.
Solving identities is quite simple. To do this you need to do identity transformations until the goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed will be solved.
You will need
Instructions
The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many trigonometric formulas, which are essentially the same identities.
Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b ^2=a^2+2ab+b^2.
Simplify both
Today we will talk about logarithmic formulas and we will give indicative solution examples.
They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithm formulas to solve, let us remind you of all the properties:
Now, based on these formulas (properties), we will show examples of solving logarithms.
Logarithm a positive number b to base a (denoted by log a b) is an exponent to which a must be raised to get b, with b > 0, a > 0, and 1.
According to the definition, log a b = x, which is equivalent to a x = b, therefore log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2, because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.
log 10 100 = 2, because 10 2 = 100
Natural logarithm- also an ordinary logarithm, a logarithm, but with the base e (e = 2.71828... - an irrational number). Denoted as ln.
It is advisable to memorize the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
log 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
Exponent of the logarithmic number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at examples of solving logarithmic equations in more detail in the article: "". Do not miss!
If you still have questions about the solution, write them in the comments to the article.
Note: we decided to get a different class of education and study abroad as an option.
274. Remarks.
A) If the expression you want to evaluate contains sum or difference numbers, then they must be found without the help of tables by ordinary addition or subtraction. Eg:
log (35 +7.24) 5 = 5 log (35 + 7.24) = 5 log 42.24.
b) Knowing how to logarithm expressions, we can, inversely, using a given logarithm result, find the expression from which this result was obtained; so if
log X=log a+ log b- 3 log With,
then it is easy to understand that
V) Before moving on to considering the structure of logarithmic tables, we will indicate some properties of decimal logarithms, i.e. those in which the number 10 is taken as the base (only such logarithms are used for calculations).
Chapter two.
Properties of decimal logarithms.
275 . A) Since 10 1 = 10, 10 2 = 100, 10 3 = 1000, 10 4 = 10000, etc., then log 10 = 1, log 100 = 2, log 1000 = 3, log 10000 = 4, and etc.
Means, The logarithm of an integer represented by one and zeros is a positive integer containing as many ones as there are zeros in the representation of the number.
Thus: log 100,000 = 5, log 1000 000 = 6 , etc.
b) Because
log 0.1 = -l; log 0.01 = - 2; log 0.001 == -3; log 0.0001 = - 4, etc.
Means, The logarithm of a decimal fraction, represented by a unit with preceding zeros, is a negative integer containing as many negative units as there are zeros in the representation of the fraction, including 0 integers.
Thus: log 0.00001= - 5, log 0.000001 = -6, etc.
V) Let's take an integer that is not represented by one and zeros, for example. 35, or a whole number with a fraction, for example. 10.7. The logarithm of such a number cannot be an integer, since raising 10 to a power with an integer exponent (positive or negative), we get 1 with zeros (following 1, or preceding it). Let us now assume that the logarithm of such a number is some fraction a / b . Then we would have equality
But these equalities are impossible, as 10A there are 1s with zeros, whereas degrees 35b And 10,7b by any measure b cannot give 1 followed by zeros. This means that we cannot allow log 35 And log 10.7 were equal to fractions. But from the properties of the logarithmic function we know () that every positive number has a logarithm; consequently, each of the numbers 35 and 10.7 has its own logarithm, and since it cannot be either an integer number or a fractional number, it is an irrational number and, therefore, cannot be expressed exactly by means of numbers. Irrational logarithms are usually expressed approximately as a decimal fraction with several decimal places. The integer number of this fraction (even if it were “0 integers”) is called characteristic, and the fractional part is the mantissa of the logarithm. If, for example, there is a logarithm 1,5441 , then its characteristic is equal 1 , and the mantissa is 0,5441 .
G) Let's take some integer or mixed number, for example. 623 or 623,57 . The logarithm of such a number consists of a characteristic and a mantissa. It turns out that decimal logarithms have the convenience that we can always find their characteristics by one type of number . To do this, we count how many digits are in a given whole number, or in an integer part mixed number, In our examples of these numbers 3 . Therefore, each of the numbers 623 And 623,57 more than 100 but less than 1000; this means that the logarithm of each of them is greater log 100, i.e. more 2 , but less log 1000, i.e. less 3 (remember that a larger number also has a larger logarithm). Hence, log 623 = 2,..., And log 623.57 = 2,... (dots replace unknown mantissas).
Like this we find:
|
10 < 56,7 < 100 1 < log56,7 < 2 log 56.7 = 1,... |
1000 < 8634 < 10 000 3 < log8634 < 4 log 8634 = 3,... |
Let in general a given integer number, or an integer part of a given mixed number, contain m numbers Since the smallest integer containing m numbers, yes 1 With m - 1 zeros at the end, then (denoting this number N) we can write the inequalities:
and therefore,
m - 1 < log N < m ,
log N = ( m- 1) + positive fraction.
So the characteristic logN = m - 1 .
We see in this way that the characteristic of the logarithm of an integer or mixed number contains as many positive units as there are digits in the integer part of the number minus one.
Having noticed this, we can directly write:
log 7.205 = 0,...; log 83 = 1,...; log 720.4 = 2,... and so on.
d) Let's take several decimal fractions smaller 1 (i.e. having 0 whole): 0,35; 0,07; 0,0056; 0,0008, and so on.
Thus, each of these logarithms is contained between two negative integers that differ by one unit; therefore each of them is equal to the smaller of these negative numbers increased by some positive fraction. For example, log0.0056= -3 + positive fraction. Let's assume that this fraction is 0.7482. Then it means:
log 0.0056 = - 3 + 0.7482 (= - 2.2518).
Amounts such as - 3 + 0,7482 , consisting of a negative integer and a positive decimal fraction, we agreed to write abbreviated as follows in logarithmic calculations: 3 ,7482 (This number reads: 3 minus, 7482 ten thousandths.), i.e. they put a minus sign over the characteristic in order to show that it relates only to this characteristic, and not to the mantissa, which remains positive. Thus, from the above table it is clear that
log 0.35 == 1 ,....; log 0.07 = 2,....; log 0.0008 = 4 ,....
Let at all 
. there is a decimal fraction in which before the first significant digit α
costs m
zeros, including 0 integers. Then it is obvious that
- m < log A < - (m- 1).
Since from two integers:- m And - (m- 1) there is less - m , That
log A = - m+ positive fraction,
and therefore the characteristic log A = - m (with a positive mantissa).
Thus, the characteristic of the logarithm of a decimal fraction less than 1 contains as many negative ones as there are zeros in the image of the decimal fraction before the first significant digit, including zero integers; The mantissa of such a logarithm is positive.
e) Let's multiply some number N(integer or fractional - it doesn’t matter) by 10, by 100 by 1000..., in general by 1 with zeros. Let's see how this changes log N. Since the logarithm of the product is equal to the sum of the logarithms of the factors, then
log(N 10) = log N + log 10 = log N + 1;
log(N 100) = log N + log 100 = log N + 2;
log(N 1000) = log N + log 1000 = log N + 3; etc.
When to log N we add some integer, then we can always add this number to the characteristic, and not to the mantissa.
So, if log N = 2.7804, then 2.7804 + 1 = 3.7804; 2.7804 + 2 = 4.7801, etc.;
or if log N = 3.5649, then 3.5649 + 1 = 2.5649; 3.5649 + 2 = 1.5649, etc.
When a number is multiplied by 10, 100, 1000,..., generally by 1 with zeros, the mantissa of the logarithm does not change, and the characteristic increases by as many units as there are zeros in the factor .
Similarly, taking into account that the logarithm of the quotient is equal to the logarithm of the dividend without the logarithm of the divisor, we get:
log N / 10 = log N- log 10 = log N -1;
log N / 100 = log N- log 100 = log N -2;
log N / 1000 = log N- log 1000 = log N -3; and so on.
If we agree, when subtracting an integer from a logarithm, to always subtract this integer from the characteristic and leave the mantissa unchanged, then we can say:
Dividing a number by 1 with zeros does not change the mantissa of the logarithm, but the characteristic decreases by as many units as there are zeros in the divisor.
276. Consequences. From property ( e) the following two corollaries can be deduced:
A) The mantissa of the logarithm of a decimal number does not change when moved to a decimal point , because moving a decimal point is equivalent to multiplying or dividing by 10, 100, 1000, etc. Thus, logarithms of numbers:
0,00423, 0,0423, 4,23, 423
differ only in characteristics, but not in mantissas (provided that all mantissas are positive).
b) The mantissas of numbers that have the same significant part, but differ only by ending zeros, are the same: Thus, the logarithms of numbers: 23, 230, 2300, 23,000 differ only in characteristics.
Comment. From the indicated properties of decimal logarithms it is clear that we can find the characteristics of the logarithm of an integer and a decimal fraction without the help of tables (this is the great convenience of decimal logarithms); as a result, only one mantissa is placed in logarithmic tables; in addition, since finding the logarithms of fractions is reduced to finding the logarithms of integers (the logarithm of a fraction = the logarithm of the numerator without the logarithm of the denominator), the mantissas of logarithms of only integers are placed in the tables.
Chapter three.
Design and use of four-digit tables.
277. Systems of logarithms. A system of logarithms is a set of logarithms calculated for a number of consecutive integers using the same base. Two systems are used: the system of ordinary or decimal logarithms, in which the number is taken as the base 10 , and the system of so-called natural logarithms, in which an irrational number is taken as the basis (for some reasons that are clear in other branches of mathematics) 2,7182818 ... For calculations, decimal logarithms are used, due to the convenience that we indicated when we listed the properties of such logarithms.
Natural logarithms are also called Neperov, named after the inventor of logarithms, a Scottish mathematician Nepera(1550-1617), and decimal logarithms - Briggs named after the professor Brigga(a contemporary and friend of Napier), who first compiled tables of these logarithms.
278. Converting a negative logarithm into one whose mantissa is positive, and the inverse transformation. We have seen that the logarithms of numbers less than 1 are negative. This means that they consist of a negative characteristic and a negative mantissa. Such logarithms can always be transformed so that their mantissa is positive, but the characteristic remains negative. To do this, it is enough to add a positive one to the mantissa, and a negative one to the characteristic (which, of course, does not change the value of the logarithm).
If, for example, we have a logarithm - 2,0873 , then you can write:
- 2,0873 = - 2 - 1 + 1 - 0,0873 = - (2 + 1) + (1 - 0,0873) = - 3 + 0,9127,
or abbreviated:
Conversely, any logarithm with a negative characteristic and a positive mantissa can be turned into a negative one. To do this, it is enough to add a negative one to the positive mantissa, and a positive one to the negative characteristic: so, you can write:
279. Description of four-digit tables. To solve most practical problems, four-digit tables are quite sufficient, the handling of which is very simple. These tables (with the inscription “logarithms” at the top) are placed at the end of this book, and a small part of them (to explain the arrangement) is printed on this page. They contain mantissas
Logarithms.
logarithms of all integers from 1 before 9999 inclusive, calculated to four decimal places, with the last of these places increased by 1 in all those cases where the 5th decimal place would be 5 or more than 5; therefore, 4-digit tables give approximate mantissas up to 1 / 2 ten-thousandth part (with a deficiency or excess).
Since we can directly characterize the logarithm of an integer or a decimal fraction, based on the properties of decimal logarithms, we must take only the mantissas from the tables; At the same time, we must remember that the position of the decimal point in a decimal number, as well as the number of zeros at the end of the number, do not affect the value of the mantissa. Therefore, when finding the mantissa for a given number, we discard the comma in this number, as well as the zeros at the end of it, if there are any, and find the mantissa of the integer formed after this. The following cases may arise.
1) An integer consists of 3 digits. For example, let’s say we need to find the mantissa of the logarithm of the number 536. The first two digits of this number, i.e. 53, are found in the tables in the first vertical column on the left (see table). Having found the number 53, we move from it along a horizontal line to the right until this line intersects with a vertical column passing through one of the numbers 0, 1, 2, 3,... 9, placed at the top (and bottom) of the table, which is 3- th digit of a given number, i.e. in our example, the number 6. At the intersection we get the mantissa 7292 (i.e. 0.7292), which belongs to the logarithm of the number 536. Similarly, for the number 508 we find the mantissa 0.7059, for the number 500 we find 0.6990, etc.
2) An integer consists of 2 or 1 digits. Then we mentally assign one or two zeros to this number and find the mantissa for the three-digit number thus formed. For example, we add one zero to the number 51, from which we get 510 and find the mantissa 7070; to the number 5 we assign 2 zeros and find the mantissa 6990, etc.
3) An integer is expressed in 4 digits. For example, you need to find the mantissa of log 5436. Then first we find in the tables, as was just indicated, the mantissa for the number represented by the first 3 digits of this number, i.e. for 543 (this mantissa will be 7348); then we move from the found mantissa along the horizontal line to the right (to the right side of the table, located behind the thick vertical line) until it intersects with the vertical column passing through one of the numbers: 1, 2 3,... 9, located at the top (and at the bottom ) of this part of the table, which represents the 4th digit of a given number, i.e., in our example, the number 6. At the intersection we find the correction (number 5), which must be mentally applied to the mantissa of 7348 in order to obtain the mantissa of the number 5436; This way we get the mantissa 0.7353.
4) An integer is expressed with 5 or more digits. Then we discard all digits except the first 4, and take an approximate four-digit number, and increase the last digit of this number by 1 in that number. case when the discarded 5th digit of the number is 5 or more than 5. So, instead of 57842 we take 5784, instead of 30257 we take 3026, instead of 583263 we take 5833, etc. For this rounded four-digit number, we find the mantissa as just explained.
Guided by these instructions, let us find, for example, the logarithms of the following numbers:
36,5; 804,7; 0,26; 0,00345; 7,2634; 3456,06.
First of all, without turning to the tables for now, we will put down only the characteristics, leaving room for the mantissas, which we will write out after:
log 36.5 = 1,.... log 0.00345 = 3,....
log 804.7 = 2,.... log 7.2634 = 0,....
log 0.26 = 1,.... log 3456.86 = 3,....
log 36.5 = 1.5623; log 0.00345 = 3.5378;
log 804.7 = 2.9057; log 7.2634 = 0.8611;
log 0.26 = 1.4150; log 3456.86 = 3.5387.
280. Note. In some four-digit tables (for example, in tables V. Lorchenko and N. Ogloblina, S. Glazenap, N. Kamenshchikova) corrections for the 4th digit of this number are not placed. When dealing with such tables, you have to find these corrections using simple calculation, which can be performed on the basis of the following truth: if the numbers exceed 100, and the differences between them are less than 1, then without sensitive error it can be accepted that differences between logarithms are proportional to differences between corresponding numbers . Let, for example, we need to find the mantissa corresponding to the number 5367. This mantissa, of course, is the same as for the number 536.7. We find in the tables for the number 536 the mantissa 7292. Comparing this mantissa with the mantissa 7300 adjacent to the right, corresponding to the number 537, we notice that if the number 536 increases by 1, then its mantissa will increase by 8 ten-thousandths (8 is the so-called table difference between two adjacent mantissas); if the number 536 increases by 0.7, then its mantissa will increase not by 8 ten-thousandths, but by some smaller number X ten thousandths, which, according to the assumed proportionality, must satisfy the proportions:
X :8 = 0.7:1; where X = 8 07 = 5,6,
which is rounded to 6 ten-thousandths. This means that the mantissa for the number 536.7 (and therefore for the number 5367) will be: 7292 + 6 = 7298.
Note that finding an intermediate number using two adjacent numbers in tables is called interpolation. The interpolation described here is called proportional, since it is based on the assumption that the change in the logarithm is proportional to the change in the number. It is also called linear, since it assumes that graphically the change in a logarithmic function is expressed by a straight line.
281. Error limit of the approximate logarithm. If the number whose logarithm is being sought is an exact number, then the limit of error of its logarithm found in 4-digit tables can, as we said in, be taken 1 / 2 ten-thousandth part. If this number is not exact, then to this error limit we must also add the limit of another error resulting from the inaccuracy of the number itself. It has been proven (we omit this proof) that such a limit can be taken to be the product
a(d +1) ten thousandths.,
in which A is the margin of error for the most imprecise number, assuming that its integer part contains 3 digits, a d tabular difference of mantissas corresponding to two consecutive three-digit numbers between which the given imprecise number lies. Thus, the limit of the final error of the logarithm will then be expressed by the formula:
1 / 2 + a(d +1) ten thousandths
Example. Find log π , taking for π approximate number 3.14, exact to 1 / 2 hundredth.
Moving the comma after the 3rd digit in the number 3.14, counting from the left, we get three digit number 314, exact to 1 / 2 units; This means that the margin of error for an inaccurate number, i.e., what we denoted by the letter A , there is 1 / 2 From the tables we find:
log 3.14 = 0.4969.
Table difference d between the mantissas of the numbers 314 and 315 is equal to 14, so the error of the found logarithm will be less
1 / 2 + 1 / 2 (14 +1) = 8 ten thousandths.
Since we do not know about the logarithm 0.4969 whether it is deficient or excessive, we can only guarantee that the exact logarithm π lies between 0.4969 - 0.0008 and 0.4969 + 0.0008, i.e. 0.4961< log π < 0,4977.
282. Find a number using a given logarithm. To find a number using a given logarithm, the same tables can be used to find the mantissas of given numbers; but it is more convenient to use other tables that contain the so-called antilogarithms, i.e., numbers corresponding to these mantissas. These tables, indicated by the inscription at the top “antilogarithms,” are placed at the end of this book after the tables of logarithms; a small part of them is placed on this page (for explanation).
Suppose you are given a 4-digit mantissa 2863 (we do not pay attention to the characteristic) and you need to find the corresponding integer. Then, having tables of antilogarithms, you need to use them in exactly the same way as was previously explained to find the mantissa for a given number, namely: we find the first 2 digits of the mantissa in the first column on the left. Then we move from these numbers along the horizontal line to the right until it intersects with the vertical column coming from the 3rd digit of the mantissa, which must be looked for in the top line (or bottom). At the intersection we find the four-digit number 1932, corresponding to the mantissa 286. Then from this number we move further along the horizontal line to the right until the intersection with the vertical column coming from the 4th digit of the mantissa, which must be found at the top (or bottom) among the numbers 1, 2 placed there , 3,... 9. At the intersection we find correction 1, which must be applied (in the mind) to the number 1032 found earlier in order to obtain the number corresponding to the mantissa 2863.
Thus, the number will be 1933. After this, paying attention to the characteristic, you need to put occupied in the proper place in the number 1933. For example:
If log x = 3.2863, then X = 1933,
„ log x = 1,2863, „ X = 19,33,
, log x = 0,2&63, „ X = 1,933,
„ log x = 2 ,2863, „ X = 0,01933
Here are more examples:
log x = 0,2287, X = 1,693,
log x = 1 ,7635, X = 0,5801,
log x = 3,5029, X = 3184,
log x = 2 ,0436, X = 0,01106.
If the mantissa contains 5 or more digits, then we take only the first 4 digits, discarding the rest (and increasing the 4th digit by 1 if the 5th digit has five or more). For example, instead of the mantissa 35478 we take 3548, instead of 47562 we take 4756.
283. Note. The correction for the 4th and subsequent digits of the mantissa can also be found through interpolation. So, if the mantissa is 84357, then, having found the number 6966, corresponding to the mantissa 843, we can further reason as follows: if the mantissa increases by 1 (thousandth), i.e., it makes 844, then the number, as can be seen from the tables, will increase by 16 units; if the mantissa increases not by 1 (thousandth), but by 0.57 (thousandth), then the number will increase by X units, and X must satisfy the proportions:
X : 16 = 0.57: 1, from where x = 16 0,57 = 9,12.
This means that the required number will be 6966+ 9.12 = 6975.12 or (limited to only four digits) 6975.
284. Error limit of the found number. It has been proven that in the case when in the found number the comma is after the 3rd digit from the left, i.e. when the characteristic of the logarithm is 2, the sum can be taken as the error limit
Where A is the error limit of the logarithm (expressed in ten thousandths) by which the number was found, and d - the difference between the mantissas of two three-digit consecutive numbers between which the found number lies (with a comma after the 3rd digit from the left). When the characteristic is not 2, but some other, then in the found number the comma will have to be moved to the left or to the right, i.e., divide or multiply the number by some power of 10. In this case, the error of the result will also be divided or multiplied by the same power of 10.
Let, for example, we are looking for a number using the logarithm 1,5950 , which is known to be accurate to 3 ten-thousandths; that means then A = 3 . The number corresponding to this logarithm, found from the table of antilogarithms, is 39,36 . Moving the comma after the 3rd digit from the left, we have the number 393,6 , consisting between 393 And 394 . From the tables of logarithms we see that the difference between the mantissas corresponding to these two numbers is 11 ten thousandths; Means d = 11 . The error of the number 393.6 will be less
This means that the error in the number 39,36 there will be less 0,05 .
285. Operations on logarithms with negative characteristics. Adding and subtracting logarithms does not present any difficulties, as can be seen from the following examples:
There is also no difficulty in multiplying the logarithm by a positive number, for example:
In the last example, the positive mantissa is separately multiplied by 34, then the negative characteristic is multiplied by 34.
If the logarithm of a negative characteristic and a positive mantissa is multiplied by a negative number, then proceed in two ways: either the given logarithm is first turned negative, or the mantissa and characteristic are multiplied separately and the results are combined together, for example:
3 ,5632 (- 4) = - 2,4368 (- 4) = 9,7472;
3 ,5632 (- 4) = + 12 - 2,2528 = 9,7472.
When dividing, two cases may arise: 1) the negative characteristic is divided and 2) is not divisible by a divisor. In the first case, the characteristic and mantissa are separated separately:
10 ,3784: 5 = 2 ,0757.
In the second case, so many negative units are added to the characteristic so that the resulting number is divided by the divisor; the same number of positive units is added to the mantissa:
3 ,7608: 8 = (- 8 + 5,7608) : 8 = 1 ,7201.
This transformation must be done in the mind, so the action goes like this:
286. Replacing subtracted logarithms with terms. When calculating some complex expression using logarithms, you have to add some logarithms and subtract others; in this case, in the usual way of performing actions, they separately find the sum of the added logarithms, then the sum of the subtracted ones, and subtract the second from the first sum. For example, if we have:
log X = 2,7305 - 2 ,0740 + 3 ,5464 - 8,3589 ,
then the usual execution of actions will look like this:
However, it is possible to replace subtraction with addition. So:
Now you can arrange the calculation like this:
287. Examples of calculations.
Example 1. Evaluate expression:
If A = 0.8216, B = 0.04826, C = 0.005127 And D = 7.246.
Let's take a logarithm of this expression:
log X= 1/3 log A + 4 log B - 3 log C - 1/3 log D
Now, to avoid unnecessary loss of time and to reduce the possibility of errors, first of all we will arrange all the calculations without executing them for now and, therefore, without referring to the tables:
After this, we take the tables and put logarithms on the remaining free places:
Error limit. First, let's find the limit of error of the number x 1 = 194,5 , equal to:
So, first of all you need to find A , i.e., the error limit of the approximate logarithm, expressed in ten thousandths. Let us assume that these numbers A, B, C And D all are accurate. Then the errors in individual logarithms will be as follows (in ten thousandths):
V logA.......... 1 / 2
V 1/3 log A......... 1 / 6 + 1 / 2 = 2 / 3
( 1 / 2 added because when dividing by 3 logarithms of 1.9146, we rounded the quotient by discarding its 5th digit, and, therefore, made an even smaller error 1 / 2 ten-thousandth).
Now we find the error limit of the logarithm:
A = 2 / 3 + 2 + 3 / 2 + 1 / 6 = 4 1 / 3 (ten thousandths).
Let us further define d . Because x 1 = 194,5 , then 2 consecutive integers between which lies x 1 will 194 And 195 . Table difference d between the mantissas corresponding to these numbers is equal to 22 . This means that the limit of error of the number is x 1 There is:
Because x = x 1 : 10, then the error limit in the number x equals 0,3:10 = 0,03 . Thus, the number we found 19,45 differs from the exact number by less than 0,03 . Since we do not know whether our approximation was found with a deficiency or with an excess, we can only guarantee that
19,45 + 0,03 > X > 19,45 - 0,03 , i.e.
19,48 > X > 19,42 ,
and therefore, if we accept X =19,4 , then we will have an approximation with a disadvantage with an accuracy of up to 0.1.
Example 2. Calculate:
X = (- 2,31) 3 5 √72 = - (2,31) 3 5 √72 .
Since negative numbers do not have logarithms, we first find:
X" = (2,31) 3 5 √72
by decomposition:
log X"= 3 log 2.31 + 1 / 5 log72.
After calculation it turns out:
X" = 28,99 ;
hence,
x = - 28,99 .
Example 3. Calculate:
Continuous logarithmization cannot be used here, since the sign of the root is c u m m a. In such cases, calculate the formula by parts.
First we find N = 5 √8 , Then N 1 = 4 √3 ; then by simple addition we determine N+ N 1 , and finally we calculate 3 √N+ N 1 ; it turns out:
N=1.514, N 1 = 1,316 ; N+ N 1 = 2,830 .
log x= log 3 √ 2,830 = 1 / 3 log 2.830 = 0,1506 ;
x = 1,415 .
Chapter Four.
Exponential and logarithmic equations.
288. Exponential equations are those in which the unknown is included in the exponent, and logarithmic- those in which the unknown enters under the sign log. Such equations can be solvable only in special cases, and one has to rely on the properties of logarithms and on the principle that if the numbers are equal, then their logarithms are equal, and, conversely, if the logarithms are equal, then the corresponding numbers are equal.
Example 1. Solve the equation: 2 x = 1024 .
Let's logarithm both sides of the equation:
Example 2. Solve the equation: a 2x - a x = 1 . Putting a x = at , we get a quadratic equation:
y 2 - at - 1 = 0 ,
Because 1-√5 < 0 , then the last equation is impossible (function a x there is always a positive number), and the first gives:
Example 3. Solve the equation:
log( a + x) + log ( b + x) = log ( c + x) .
The equation can be written like this:
log[( a + x) (b + x)] = log ( c + x) .
From the equality of logarithms we conclude that the numbers are equal:
(a + x) (b + x) = c + x .
This is a quadratic equation, the solution of which is not difficult.
Chapter five.
Compound interest, term payments and term payments.
289. Basic problem on compound interest. How much will the capital turn into? A rubles, given in growth at R compound interest, after the lapse of t years ( t - integer)?
They say that capital is paid at compound interest if the so-called “interest on interest” is taken into account, that is, if the interest money due on the capital is added to the capital at the end of each year in order to increase it with interest in subsequent years.
Every ruble of capital given away R %, will bring profit within one year p / 100 ruble, and, therefore, every ruble of capital in 1 year will turn into 1 + p / 100 ruble (for example, if capital is given at 5 %, then every ruble of it in a year will turn into 1 + 5 / 100 , i.e. in 1,05 ruble).
For brevity, denoting the fraction p / 100 with one letter, for example, r , we can say that every ruble of capital in a year will turn into 1 + r rubles; hence, A rubles will be returned in 1 year to A (1 + r ) rub. After another year, i.e. 2 years from the start of growth, every ruble of these A (1 + r ) rub. will contact again 1 + r rub.; This means that all capital will turn into A (1 + r ) 2 rub. In the same way we find that after three years the capital will be A (1 + r ) 3 , in four years it will be A (1 + r ) 4 ,... generally through t years if t is an integer, it will turn to A (1 + r ) t rub. Thus, denoting by A final capital, we will have the following compound interest formula:
A = A (1 + r ) t Where r = p / 100 .
Example. Let a =2,300 rub., p = 4, t=20 years; then the formula gives:
r = 4 / 100 = 0,04 ; A = 2,300 (1.04) 20.
To calculate A, we use logarithms:
log a = log 2 300 + 20 log 1.04 = 3.3617 + 20 0.0170 = 3.3617+0.3400 = 3.7017.
A = 5031 ruble.
Comment. In this example we had to log 1.04 multiply by 20 . Since the number 0,0170 there is an approximate value log 1.04 up to 1 / 2 ten-thousandth part, then the product of this number by 20 it will definitely only be until 1 / 2 20, i.e. up to 10 ten-thousandths = 1 thousandth. Therefore in total 3,7017 We cannot vouch not only for the number of ten thousandths, but also for the number of thousandths. In order to obtain greater accuracy in such cases, it is better for the number 1 + r take logarithms not 4-digit, but with a large number numbers, eg. 7-digit. For this purpose, we present here a small table in which 7-digit logarithms are written out for the most common values R .
290. The main task is for urgent payments. Someone took A rubles per R % with the condition to repay the debt, together with the interest due on it, in t years, paying the same amount at the end of each year. What should this amount be?
Sum x , paid annually under such conditions, is called urgent payment. Let us again denote by the letter r annual interest money from 1 rub., i.e. the number p / 100 . Then by the end of the first year the debt A increases to A (1 + r ), basic payment X it will cost rubles A (1 + r )-X .
By the end of the second year, every ruble of this amount will again turn into 1 + r rubles, and therefore the debt will be [ A (1 + r )-X ](1 + r ) = A (1 + r ) 2 - x (1 + r ), and for payment x rubles will be: A (1 + r ) 2 - x (1 + r ) - X . In the same way, we will make sure that by the end of the 3rd year the debt will be
A (1 + r ) 3 - x (1 + r ) 2 - x (1 + r ) - x ,
and in general and the end t year it will be:
A (1 + r ) t - x (1 + r ) t -1 - x (1 + r ) t -2 ... - x (1 + r ) - x , or
A (1 + r ) t - x [ 1 + (1 + r ) + (1 + r ) 2 + ...+ (1 + r ) t -2 + (1 + r ) t -1 ]
The polynomial inside the parentheses represents the sum of the terms of a geometric progression; which has the first member 1 , last ( 1 + r ) t -1, and the denominator ( 1 + r ). Using the formula for the sum of terms of a geometric progression (Section 10, Chapter 3, § 249), we find:
and the amount of debt after t -th payment will be:
According to the conditions of the problem, the debt is at the end t -th year must be equal to 0 ; That's why:
where
When calculating this urgent payment formulas using logarithms we must first find the auxiliary number N = (1 + r ) t by logarithm: log N= t log(1+ r) ; having found N, subtract 1 from it, then we get the denominator of the formula for X, after which we find by secondary logarithm:
log X=log a+ log N + log r - log (N - 1).
291. The main task for term contributions. Someone deposits the same amount into the bank at the beginning of each year. A rub. Determine what capital will be formed from these contributions after t years if the bank pays R compound interest.
Designated by r annual interest money from 1 ruble, i.e. p / 100 , we reason like this: by the end of the first year the capital will be A (1 + r );
at the beginning of the 2nd year will be added to this amount A rubles; this means that at this time capital will be A (1 + r ) + a . By the end of the 2nd year he will be A (1 + r ) 2 + a (1 + r );
at the beginning of the 3rd year it is entered again A rubles; this means that at this time there will be capital A (1 + r ) 2 + a (1 + r ) + A ; by the end of the 3rd he will be A (1 + r ) 3 + a (1 + r ) 2 + a (1 + r ) Continuing these arguments further, we find that by the end t year the required capital A will:
This is the formula for term contributions made at the beginning of each year.
The same formula can be obtained by the following reasoning: down payment to A rubles while in the bank t years, will turn, according to the compound interest formula, into A (1 + r ) t rub. The second installment, being in the bank for one year less, i.e. t - 1 years old, contact A (1 + r ) t- 1 rub. Likewise, the third installment will give A (1 + r ) t-2 etc., and finally the last installment, having been in the bank for only 1 year, will go to A (1 + r ) rub. This means the final capital A rub. will:
A= A (1 + r ) t + A (1 + r ) t- 1 + A (1 + r ) t-2 + . . . + A (1 + r ),
which, after simplification, gives the formula found above.
When calculating using logarithms of this formula, you must proceed in the same way as when calculating the formula for urgent payments, i.e., first find the number N = ( 1 + r ) t by its logarithm: log N= t log(1 + r ), then the number N- 1 and then take a logarithm of the formula:
log A = log a+log(1+ r) + log (N - 1) - 1ogr
Comment. If an urgent contribution to A rub. was made not at the beginning, but at the end of each year (as, for example, an urgent payment is made X to pay off the debt), then, reasoning similarly to the previous one, we find that by the end t year the required capital A" rub. will be (including the last installment A rub., not bearing interest):
A"= A (1 + r ) t- 1 + A (1 + r ) t-2 + . . . + A (1 + r ) + A
which is equal to:
i.e. A" ends up in ( 1 + r ) times less A, which was to be expected, since every ruble of capital A" lies in the bank for a year less than the corresponding ruble of capital A.
Logarithm of the number b (b > 0) to base a (a > 0, a ≠ 1)– exponent to which the number a must be raised to obtain b.
The base 10 logarithm of b can be written as log(b), and the logarithm to base e (natural logarithm) is ln(b).
Often used when solving problems with logarithms:
There are four main properties of logarithms.
Let a > 0, a ≠ 1, x > 0 and y > 0.
Logarithm of the product equal to the sum of logarithms:
log a (x ⋅ y) = log a x + log a y
Logarithm of the quotient equal to the difference of logarithms:
log a (x / y) = log a x – log a y
Logarithm of degree equal to the product powers per logarithm:
If the base of the logarithm is in the degree, then another formula applies:
This property can be obtained from the property of the logarithm of a power, since the root of the nth power equal to the power 1/n:
This formula is also often used to solve various tasks to logarithms:
Special case:
Let us have 2 functions f(x) and g(x) under logarithms with the same bases and between them there is an inequality sign:
To compare them, you need to first look at the base of the logarithms a:
Problems with logarithms included in the Unified State Examination in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the appropriate sections. Also, tasks with logarithms are found in the math task bank. You can find all examples by searching the site.
Logarithms have always been considered a difficult topic in school mathematics courses. There are many different definitions of logarithm, but for some reason most textbooks use the most complex and unsuccessful of them.
We will define the logarithm simply and clearly. To do this, let's create a table:
So, we have powers of two.
If you take the number from the bottom line, you can easily find the power to which you will have to raise two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.
And now - actually, the definition of the logarithm:
the base a of the argument x is the power to which the number a must be raised to obtain the number x.
Designation: log a x = b, where a is the base, x is the argument, b is what the logarithm is actually equal to.
For example, 2 3 = 8 ⇒log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success, log 2 64 = 6, since 2 6 = 64.
The operation of finding the logarithm of a number to a given base is called. So, let's add a new line to our table:
| 2 1 | 2 2 | 2 3 | 2 4 | 2 5 | 2 6 |
| 2 | 4 | 8 | 16 | 32 | 64 |
| log 2 2 = 1 | log 2 4 = 2 | log 2 8 = 3 | log 2 16 = 4 | log 2 32 = 5 | log 2 64 = 6 |
Unfortunately, not all logarithms are calculated so easily. For example, try to find log 2 5. The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the interval. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.
Such numbers are called irrational: the numbers after the decimal point can be written ad infinitum, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.
It is important to understand that a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:
Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.
We've figured out the definition - all that remains is to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:
Such restrictions are called range of acceptable values(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒x > 0, a > 0, a ≠ 1.
Note that there are no restrictions on the number b (the value of the logarithm). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.
However, now we are considering only numerical expressions, where it is not required to know the VA of the logarithm. All restrictions have already been taken into account by the authors of the problems. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain very strong constructions that do not necessarily correspond to the above restrictions.
Now let's consider general scheme calculating logarithms. It consists of three steps:
That's all! If the logarithm turns out to be irrational, this will be visible already in the first step. The requirement that the base be greater than one is very important: this reduces the likelihood of error and greatly simplifies the calculations. Same with decimals: if you immediately convert them to regular ones, there will be many fewer errors.
Let's see how this scheme works using specific examples:
Task. Calculate the logarithm: log 5 25
Let's create and solve the equation:
log 5 25 = b ⇒(5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2;
Task. Calculate the logarithm:
Task. Calculate the logarithm: log 4 64
Task. Calculate the logarithm: log 16 1
Task. Calculate the logarithm: log 7 14
A small note on the last example. How can you be sure that a number is not an exact power of another number? It’s very simple - just factor it into prime factors. If the expansion has at least two different factors, the number is not an exact power.
Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14.
8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;
Let us also note that we ourselves prime numbers are always exact degrees of themselves.
Some logarithms are so common that they have a special name and symbol.
of the argument x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain the number x. Designation: lg x.
For example, log 10 = 1; lg 100 = 2; lg 1000 = 3 - etc.
From now on, when a phrase like “Find lg 0.01” appears in a textbook, know that this is not a typo. This is a decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x
Everything that is true for ordinary logarithms is also true for decimal logarithms.
There is another logarithm that has its own designation. In some ways, it's even more important than decimal. It's about about the natural logarithm.
of the argument x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x.
Many people will ask: what is the number e? This is an irrational number, its exact value impossible to find and record. I will give only the first figures:
e = 2.718281828459…
We will not go into detail about what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x
Thus ln e = 1; ln e 2 = 2; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, for one: ln 1 = 0.
For natural logarithms, all the rules that are true for ordinary logarithms are valid.
See also:
How to represent a number as a logarithm?
We use the definition of logarithm.
A logarithm is an exponent to which the base must be raised to obtain the number under the logarithm sign.
Thus, in order to represent a certain number c as a logarithm to base a, you need to put a power with the same base as the base of the logarithm under the sign of the logarithm, and write this number c as the exponent:
Absolutely any number can be represented as a logarithm - positive, negative, integer, fractional, rational, irrational:
In order not to confuse a and c under stressful conditions of a test or exam, you can use the following memorization rule:
what is below goes down, what is above goes up.
For example, you need to represent the number 2 as a logarithm to base 3.
We have two numbers - 2 and 3. These numbers are the base and the exponent, which we will write under the sign of the logarithm. It remains to determine which of these numbers should be written down, to the base of the degree, and which – up, to the exponent.
The base 3 in the notation of a logarithm is at the bottom, which means that when we represent two as a logarithm to the base 3, we will also write 3 down to the base.
2 is higher than three. And in notation of the degree two we write above the three, that is, as an exponent:
Logarithm positive number b based on a, Where a > 0, a ≠ 1, is called the exponent to which the number must be raised a, To obtain b.
Definition of logarithm can be briefly written like this:
This equality is valid for b > 0, a > 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called by logarithm.
Properties of logarithms:
Logarithm of the product:
Logarithm of the quotient:
Replacing the logarithm base:
Logarithm of degree:
Logarithm of the root:
Logarithm with power base:
Decimal logarithm numbers call the logarithm of this number to base 10 and write lg b
Natural logarithm numbers are called the logarithm of that number to the base e, Where e- an irrational number approximately equal to 2.7. At the same time they write ln b.
Other notes on algebra and geometry
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.
You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Note: key moment Here - identical grounds. If the reasons are different, these rules do not work!
These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see the lesson “What is a logarithm”). Take a look at the examples and see:
Log 6 4 + log 6 9.
Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 − log 2 3.
The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 − log 3 5.
Again the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.
As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact test papers. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself.
This is what is most often required.
Task. Find the value of the expression: log 7 49 6 .
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:
I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm log a x be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we set c = x, we get:
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in conventional numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:
Task. Find the value of the expression: log 5 16 log 2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let’s “reverse” the second logarithm:
Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Often in the solution process it is necessary to represent a number as a logarithm to a given base.
In this case, the following formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.
The second formula is actually a paraphrased definition. That's what it's called: .
In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the same number a. Read this paragraph carefully again - many people get stuck on it.
Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
Note that log 25 64 = log 5 8 - simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:
If anyone doesn’t know, this was a real task from the Unified State Exam :)
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.
A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.
For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate types of logarithmic expressions:
Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values of logarithms, you should remember their properties and the sequence of actions when solving them.
In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the even root of negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:
For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.
Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.
To accurately determine the value of an unknown degree, you need to learn how to work with a table of degrees. It looks like this:
As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However for large values you will need a table of degrees. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!
It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
Given an expression of the following form: log 2 (x-1) > 3 - it is logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.
The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific numerical values in the answer, while when solving an inequality, both the range of acceptable values and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.
When solving primitive tasks of finding the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.
This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.
Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;
but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.
The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, but certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or lead to general appearance. Simplify long ones logarithmic expressions possible if you use their properties correctly. Let's get to know them quickly.
When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.
Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.
So, let's look at examples of using the basic theorems about logarithms.
Logarithms are often found in entrance exams, especially many logarithmic problems in the Unified State Exam (state exam for all school graduates). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.
Examples and solutions to problems are taken from official Unified State Exam options. Let's see how such tasks are solved.
Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.
