Instructions
Write down 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 square root sign, 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
One of the elements of primitive level algebra is the logarithm. The name comes from Greek language from the word “number” or “power” and means the degree to which the number in the base must be raised to find the final number.
The logarithm of b to base a is an exponent that requires b to be raised to base a. The result obtained is pronounced like this: “logarithm of b to base a.” The solution to logarithmic problems is that you need to determine the given power in numbers from the specified numbers. There are some basic rules to determine or solve the logarithm, as well as convert the notation itself. Using them, the solution is made logarithmic equations, derivatives are found, integrals are solved, and many other operations are performed. Basically, the solution to the logarithm itself is its simplified notation. Below are the basic formulas and properties:
For any a ; a > 0; a ≠ 1 and for any x ; y > 0.
Please note: if the base logarithm is 10, then the entry is shortened, resulting in a decimal logarithm. If there is a natural number e, then we write it down, reducing it to a natural logarithm. This means that the result of all logarithms is the power to which the base number is raised to obtain the number b.
Directly, the solution lies in calculating this degree. Before solving an expression with a logarithm, it must be simplified according to the rule, that is, using formulas. You can find the main identities by going back a little in the article.
When adding and subtracting logarithms with two different numbers but with the same bases, replace with one logarithm with the product or division of the numbers b and c, respectively. In this case, you can apply the formula for moving to another base (see above).
If you use expressions to simplify a logarithm, there are some limitations to consider. And that is: the base of the logarithm a is only a positive number, but not equal to one. The number b, like a, must be greater than zero.
There are cases where, by simplifying an expression, you will not be able to calculate the logarithm numerically. It happens that such an expression does not make sense, because many powers are irrational numbers. Under this condition, leave the power of the number as a logarithm.
\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
Let's explain it more simply. For example, \(\log_(2)(8)\) equal to the power, to which \(2\) must be raised to obtain \(8\). From this it is clear that \(\log_(2)(8)=3\).
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Examples: |
\(\log_(5)(25)=2\) |
because \(5^(2)=25\) |
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\(\log_(3)(81)=4\) |
because \(3^(4)=81\) |
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\(\log_(2)\)\(\frac(1)(32)\) \(=-5\) |
because \(2^(-5)=\)\(\frac(1)(32)\) |
Any logarithm has the following “anatomy”:
The argument of a logarithm is usually written at its level, and the base is written in subscript closer to the logarithm sign. And this entry reads like this: “logarithm of twenty-five to base five.”
For example, calculate the logarithm: a) \(\log_(4)(16)\) b) \(\log_(3)\)\(\frac(1)(3)\) c) \(\log_(\sqrt (5))(1)\) d) \(\log_(\sqrt(7))(\sqrt(7))\) e) \(\log_(3)(\sqrt(3))\)
a) To what power must \(4\) be raised to get \(16\)? Obviously the second one. That's why:
\(\log_(4)(16)=2\)
\(\log_(3)\)\(\frac(1)(3)\) \(=-1\)
c) To what power must \(\sqrt(5)\) be raised to get \(1\)? What power makes any number one? Zero, of course!
\(\log_(\sqrt(5))(1)=0\)
d) To what power must \(\sqrt(7)\) be raised to obtain \(\sqrt(7)\)? Firstly, any number to the first power is equal to itself.
\(\log_(\sqrt(7))(\sqrt(7))=1\)
e) To what power must \(3\) be raised to obtain \(\sqrt(3)\)? From we know that is a fractional power, which means Square root is the power of \(\frac(1)(2)\) .
\(\log_(3)(\sqrt(3))=\)\(\frac(1)(2)\)
Example : Calculate logarithm \(\log_(4\sqrt(2))(8)\)
Solution :
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\(\log_(4\sqrt(2))(8)=x\) |
We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of a logarithm: |
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\((4\sqrt(2))^(x)=8\) |
What connects \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: |
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\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\) |
On the left we use the properties of the degree: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\) |
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\(2^(\frac(5)(2)x)=2^(3)\) |
The bases are equal, we move on to equality of indicators |
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\(\frac(5x)(2)\) \(=3\) |
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Multiply both sides of the equation by \(\frac(2)(5)\) |
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The resulting root is the value of the logarithm |
Answer : \(\log_(4\sqrt(2))(8)=1,2\)
To understand this, let's solve the equation: \(3^(x)=9\). Just match \(x\) to make the equality work. Of course, \(x=2\).
Now solve the equation: \(3^(x)=8\).What is x equal to? That's the point.
The smartest ones will say: “X is a little less than two.” How exactly to write this number? To answer this question, the logarithm was invented. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).
I want to emphasize that \(\log_(3)(8)\), like any logarithm is just a number. Yes, it looks unusual, but it’s short. Because if we wanted to write it in the form decimal, then it would look like this: \(1.892789260714.....\)
Example : Solve the equation \(4^(5x-4)=10\)
Solution :
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\(4^(5x-4)=10\) |
\(4^(5x-4)\) and \(10\) cannot be brought to the same base. This means you can’t do without a logarithm. Let's use the definition of logarithm: |
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\(\log_(4)(10)=5x-4\) |
Let's flip the equation so that X is on the left |
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\(5x-4=\log_(4)(10)\) |
Before us. Let's move \(4\) to the right. And don't be afraid of the logarithm, treat it like an ordinary number. |
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\(5x=\log_(4)(10)+4\) |
Divide the equation by 5 |
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\(x=\)\(\frac(\log_(4)(10)+4)(5)\) |
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This is our root. Yes, it looks unusual, but they don’t choose the answer. |
Answer : \(\frac(\log_(4)(10)+4)(5)\)
As stated in the definition of a logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all the possible bases, there are two that occur so often that a special short notation was invented for logarithms with them:
That is, \(\ln(a)\) is the same as \(\log_(e)(a)\)
That is, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.
Logarithms have many properties. One of them is called the “Basic Logarithmic Identity” and looks like this:
| \(a^(\log_(a)(c))=c\) |
This property follows directly from the definition. Let's see exactly how this formula came about.
Let us recall a short notation of the definition of logarithm:
if \(a^(b)=c\), then \(\log_(a)(c)=b\)
That is, \(b\) is the same as \(\log_(a)(c)\). Then we can write \(\log_(a)(c)\) instead of \(b\) in the formula \(a^(b)=c\). It turned out \(a^(\log_(a)(c))=c\) - the main logarithmic identity.
You can find other properties of logarithms. With their help, you can simplify and calculate the values of expressions with logarithms, which are difficult to calculate directly.
Example : Find the value of the expression \(36^(\log_(6)(5))\)
Solution :
Answer : \(25\)
As mentioned above, any logarithm is just a number. The converse is also true: any number can be written as a logarithm. For example, we know that \(\log_(2)(4)\) is equal to two. Then instead of two you can write \(\log_(2)(4)\).
But \(\log_(3)(9)\) is also equal to \(2\), which means we can also write \(2=\log_(3)(9)\) . Likewise with \(\log_(5)(25)\), and with \(\log_(9)(81)\), etc. That is, it turns out
\(2=\log_(2)(4)=\log_(3)(9)=\log_(4)(16)=\log_(5)(25)=\log_(6)(36)=\ log_(7)(49)...\)
Thus, if we need, we can write two as a logarithm with any base anywhere (be it in an equation, in an expression, or in an inequality) - we simply write the base squared as an argument.
It’s the same with the triple – it can be written as \(\log_(2)(8)\), or as \(\log_(3)(27)\), or as \(\log_(4)(64) \)... Here we write the base in the cube as an argument:
\(3=\log_(2)(8)=\log_(3)(27)=\log_(4)(64)=\log_(5)(125)=\log_(6)(216)=\ log_(7)(343)...\)
And with four:
\(4=\log_(2)(16)=\log_(3)(81)=\log_(4)(256)=\log_(5)(625)=\log_(6)(1296)=\ log_(7)(2401)...\)
And with minus one:
\(-1=\) \(\log_(2)\)\(\frac(1)(2)\) \(=\) \(\log_(3)\)\(\frac(1)( 3)\) \(=\) \(\log_(4)\)\(\frac(1)(4)\) \(=\) \(\log_(5)\)\(\frac(1 )(5)\) \(=\) \(\log_(6)\)\(\frac(1)(6)\) \(=\) \(\log_(7)\)\(\frac (1)(7)\) \(...\)
And with one third:
\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)
Example : Find the meaning of the expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)
Solution :
Answer : \(1\)
The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be a positive number that is not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is equal to 2.
It is important that the scope of definition of the right and left sides of this formula is different. The left side is defined only for b>0, a>0 and a ≠ 1. The right side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the zero power, we get one.
Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against thoughtlessly using these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A similar problem exists for formula (6).
And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.
That rare case when the ODZ does not change during transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.
If we choose the number b as the new base c, we obtain an important special case of formula (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.
Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, power series expansion and representation of the function ln x using complex numbers are given.
Natural logarithm is the function y = ln x, the inverse of the exponential, x = e y, and is the logarithm to the base of the number e: ln x = log e x.
The natural logarithm is widely used in mathematics because its derivative has the simplest form: (ln x)′ = 1/ x.
Based definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
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Graph of the function y = ln x.
Graph of natural logarithm (functions y = ln x) is obtained from the exponential graph by mirror reflection relative to the straight line y = x.
The natural logarithm is defined for positive values of the variable x. It increases monotonically in its domain of definition.
At x → 0 the limit of the natural logarithm is minus infinity (-∞).
As x → + ∞, the limit of the natural logarithm is plus infinity (+ ∞). For large x, the logarithm increases quite slowly. Any power function x a with a positive exponent a grows faster than the logarithm.
The natural logarithm is a monotonically increasing function, so it has no extrema. The main properties of the natural logarithm are presented in the table.
ln 1 = 0
Formulas following from the definition of the inverse function:
Any logarithm can be expressed in terms of natural logarithms using the base substitution formula:
Proofs of these formulas are presented in the section "Logarithm".
The inverse of the natural logarithm is the exponent.
If , then
If, then.
Derivative of the natural logarithm:
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Derivative of the natural logarithm of modulus x:
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Derivative of nth order:
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Deriving formulas > > >
The integral is calculated by integration by parts:
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So,
Consider the function of the complex variable z:
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Let's express the complex variable z via module r and argument φ
:
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Using the properties of the logarithm, we have:
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Or
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The argument φ is not uniquely defined. If you put
, where n is an integer,
it will be the same number for different n.
Therefore, the natural logarithm, as a function of a complex variable, is not a single-valued function.
When the expansion takes place:
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
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