9 formulas related to the properties of powers of logarithms. Basic logarithmic identity

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.

Definition in mathematics

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.

Types of logarithms

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 individual species logarithmic expressions:

Natural logarithm ln a, where the base is the Euler number (e = 2.7).
Decimal a, where the base is 10.
Logarithm of any number b to base a>1.

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. For getting correct values logarithms, you should remember their properties and the sequence of actions when solving them.

Rules and some restrictions

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 an even root from negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:

The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
if a > 0, then a b >0, it turns out that “c” must also be greater than zero.

How to solve logarithms?

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!

Equations and inequalities

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.

The following expression is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value “x” is under the logarithmic sign. 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.

Basic theorems about logarithms

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.

The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, the mandatory condition is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.

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.

Examples of problems and inequalities

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, however, it can be applied to every mathematical inequality or logarithmic equation certain rules. First of all, you should find out whether the expression can be simplified or lead to general appearance. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them quickly.

When deciding logarithmic equations, we should 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.

How to Use Logarithm Formulas: With Examples and Solutions

So, let's look at examples of using the basic theorems about logarithms.

The property of the logarithm of a product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values out of the sign of the logarithm.

Assignments from the Unified State Exam

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 the official versions of the Unified State Exam. 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.

It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.

We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values of logarithms are found using their properties. After this, we will focus on calculating logarithms through the initially specified values of other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.

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Calculating logarithms by definition

In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.

Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.

So, calculating the logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.

Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show solutions to examples.

Example.

Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.

Solution.

The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to base 2 to the −3 power.

Similarly, we find the second logarithm: lne 5.3 =5.3.

Answer:

log 2 2 −3 =−3 and lne 5,3 =5,3.

If the number b under the logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the logarithm sign is equal to the base to the power of 1, or 2, or 3, ...

Example.

Calculate the logarithms log 5 25 , and .

Solution.

It is easy to see that 25=5 2, this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2.

Let's move on to calculating the second logarithm. The number can be represented as a power of 7: (see if necessary). Hence, .

Let's rewrite the third logarithm in the following form. Now you can see that , from which we conclude that . Therefore, by the definition of logarithm .

Briefly, the solution could be written as follows: .

Answer:

log 5 25=2 , And .

When under the logarithm sign there is a sufficiently large natural number, then it wouldn’t hurt to factor it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore calculate this logarithm by definition.

Example.

Find the value of the logarithm.

Solution.

Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of a unit and the property of the logarithm of a number, equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1 . That is, when under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.

Example.

What are logarithms and log10 equal to?

Solution.

Since , then from the definition of logarithm it follows .

In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten is equal to one, that is, lg10=lg10 1 =1.

Answer:

AND lg10=1 .

Note that the calculation of logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p, which is one of the properties of logarithms.

In practice, when a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula , which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.

Example.

Calculate the logarithm.

Solution.

Answer:

Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.

Finding logarithms through other known logarithms

The information in this paragraph continues the topic of using the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's give an example for clarification. Let's say we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .

In the above example, it was enough for us to use the property of the logarithm of a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.

Example.

Calculate the logarithm of 27 to base 60 if you know that log 60 2=a and log 60 5=b.

Solution.

So we need to find log 60 27 . It is easy to see that 27 = 3 3 , and the original logarithm, due to the property of the logarithm of the power, can be rewritten as 3·log 60 3 .

Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us to write the equality log 60 60=1. On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2·log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2·log 60 2−log 60 5=1−2·a−b.

Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3·(1−2·a−b)=3−6·a−3·b.

Answer:

log 60 27=3·(1−2·a−b)=3−6·a−3·b.

Separately, it is worth mentioning the meaning of the formula for transition to a new base of the logarithm of the form . It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values to be calculated with a certain degree of accuracy. In the next paragraph we will show how this is done.

Logarithm tables and their uses

For approximate calculation of logarithm values can be used logarithm tables. The most commonly used base 2 logarithm table, natural logarithm table, and decimal logarithm table. When working in the decimal number system, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values of logarithms.

The presented table allows you to find the values of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms using a specific example - it’s clearer this way. Let's find log1.256.

In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (digit 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the logarithm table at the intersection of the marked row and marked columns (these numbers are highlighted in orange). The sum of the marked numbers gives the desired value of the decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.

Is it possible, using the table above, to find the values of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range from 1 to 9.999? Yes, you can. Let's show how this is done with an example.

Let's calculate lg102.76332. First you need to write down number in standard form: 102.76332=1.0276332·10 2. After this, the mantissa should be rounded to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.

In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.

For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, .

Bibliography.

Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).

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 given complex function, 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 in given point y"(1)=8*e^0=8

Video on the topic

Helpful advice

Learn the table of elementary derivatives. This will significantly save time.

Sources:

derivative of a constant

So, what is the difference between rational equation from the rational? 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, in 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

- paper;
- pen.

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

General principles of the solution

Repeat from a textbook on mathematical analysis or higher mathematics what a definite integral is. As is known, the solution definite integral there is a function whose derivative gives an integrand. This function is called an antiderivative. Based on this principle, the main integrals are constructed.
Determine by the type of the integrand which of the table integrals is suitable in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.

Variable Replacement Method

If the integrand is a trigonometric function whose argument is a polynomial, then try using the change of variables method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.

Solving integrals of the second kind

If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows us to move from the rotor flux of a certain vector function to the triple integral over the divergence of a given vector field.

Substitution of integration limits

After finding the antiderivative, it is necessary to substitute the limits of integration. First, substitute the value of the upper limit into the expression for the antiderivative. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit into the antiderivative. If one of the limits of integration is infinity, then when substituting it into antiderivative function it is necessary to go to the limit and find what the expression strives for.
If the integral is two-dimensional or three-dimensional, then you will have to represent the limits of integration geometrically to understand how to evaluate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume being integrated.

The concept of logarithm and the basic logarithmic identity

The concept of logarithm and the basic logarithmic identity are closely related, because definition of logarithm in mathematical notation and is .

The basic logarithmic identity follows from the definition of the logarithm:

Definition 1

Logarithm they call the exponent $n$, when raised to which the numbers $a$ get the number $b$.

Note 1

Exponential equation$a^n=b$ for $a > 0$, $a \ne 1$ has no solutions for non-positive $b$ and has a single root for positive $b$. This root is called logarithm of the number $b$ to base $a$ and write down:

$a^(\log_(a) b)=b$.

Definition 2

Expression

$a^(\log_(a) b)=b$

called basic logarithmic identity provided that $a,b > 0$, $a \ne 1$.

Example 1

$17^(\log_(17) 6)=6$;

$e^(\ln⁡13) =13$;

$10^(\lg23)=23$.

Basic logarithmic identity

Main the logarithmic identity is called because it is used almost always when working with logarithms. In addition, with its help the basic properties of logarithms are substantiated.

Example 2

$7^5=16,807$, therefore $\log_(7)16,807=5$.

$3^(-5)=\frac(1)(243)$, therefore $\log_(3)\frac(1)(243)=-5$.

$11^0=1$, therefore $\log_(11)⁡1=0$.

Let's consider a consequence of the basic logarithmic identity:

Definition 3

If two logarithms with on the same grounds are equal, which means the logarithmic expressions are equal:

if $\log_(a)⁡b=\log_(a)⁡c$, then $b=c$.

Let's consider restrictions, which are used for the logarithmic identity:

Because when raising unity to any power we always get one, and the equality $x=\log_(a)⁡b$ exists only for $b=1$, then $\log_(1)⁡1$ will be any real number. To avoid this ambiguity, take $a \ne 1$.

The logarithm for $a=0$, according to definition, can exist only for $b=0$. Because When we raise zero to any power we always get zero, then $\log_(0)⁡0$ can be any real number. To avoid this ambiguity, take $a \ne 0$. For $a rational and irrational logarithm values, because a degree with a rational and irrational exponent can only be calculated for positive bases. To prevent this situation, take $a > 0$.

$b > 0$ follows from the condition $a > 0$, since $x=\log_(a)⁡b$, and the power of a positive number a will always be positive.

The basic logarithmic identity is often used to simplify logarithmic expressions.

Example 3

Calculate $81^(\log_(9) 7)$.

Solution.

In order to use the basic logarithmic identity, it is necessary that the base of the logarithm and the powers be the same. Let us write the base of the degree in the form:

Now we can write:

$81^(\log_(9)7)=(9^2)^(\log_(9)7)=$

Let's use the power property:

$=9^(2 \cdot \log_(9)7)=9^(\log_(9)7) \cdot 9^(\log_(9)7)=$

the basic logarithmic identity can now be applied to each factor:

$=7 \cdot 7=49$.

Note 2

To apply the basic logarithmic identity, you can also resort to replacing the base of the logarithm with the expression that appears under the logarithm sign, and vice versa.

Example 4

Calculate $7^(\frac(1)(\log_(11) 7))$.

Solution.

$7^(\frac(1)(\log_(11) 7))=7^(\log_(7) 11)=11$.

Answer: $11$.

Example 5

Calculate $7^(\frac(3)(\log_(11) 7))$.

Follows from its definition. And so the logarithm of the number b based on A is defined as the exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).

From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation a x =b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b based on a equals With. It is also clear that the topic of logarithms is closely related to the topic of powers of a number.

With logarithms, as with any numbers, you can do operations of addition, subtraction and transform in every possible way. But due to the fact that logarithms are not entirely ordinary numbers, their own special rules apply here, which are called main properties.

Adding and subtracting logarithms.

Let's take two logarithms with the same bases: log a x And log a y. Then it is possible to perform addition and subtraction operations:

log a x+ log a y= log a (x·y);

log a x - log a y = log a (x:y).

log a(x 1 . x 2 . x 3 ... x k) = log a x 1 + log a x 2 + log a x 3 + ... + log a x k.

From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore

log a 1 /b=log a 1 - log a b= - log a b.

This means there is an equality:

log a 1 / b = - log a b.

Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:

Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.