The logarithm of 12 to base 2 is equal to. Properties of logarithms and examples of their solutions

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 separate types of 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. To obtain the correct values of 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 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:

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 larger values you will need a power table. 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, 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 reduced to a general form. You can simplify long logarithmic expressions 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.

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.

Page navigation.

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 a 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 there is a sufficiently large natural number under the logarithm sign, it does not 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 one 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 logarithm sign 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 the 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 of 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).

log a r b r =log a b or log a b= log a r b r

The value of a logarithm will not change if the base of the logarithm and the number under the logarithm sign are raised to the same power.

Only positive numbers can be under the logarithm sign, and the base of the logarithm is not equal to one.

Examples.

1) Compare log 3 9 and log 9 81.

log 3 9=2, since 3 2 =9;

log 9 81=2, since 9 2 =81.

So log 3 9=log 9 81.

Note that the base of the second logarithm is equal to the square of the base of the first logarithm: 9=3 2, and the number under the sign of the second logarithm is equal to the square of the number under the sign of the first logarithm: 81=9 2. It turns out that both the number and the base of the first logarithm log 3 9 were raised to the second power, and the value of the logarithm did not change from this:

Next, since extracting the root n th degree from among A is the raising of a number A to the degree ( 1/n), then from log 9 81 you can get log 3 9 by taking the square root of the number and the base of the logarithm:

2) Check equality: log 4 25=log 0.5 0.2.

Let's look at the first logarithm. Taking the square root of the base 4 and from among 25 ; we get: log 4 25=log 2 5.

Let's look at the second logarithm. Logarithm base: 0.5= 1 / 2. The number under the sign of this logarithm: 0.2= 1/5. Let's raise each of these numbers to the minus first power:

0,5 -1 =(1 / 2) -1 =2;

0,2 -1 =(1 / 5) -1 =5.

So log 0.5 0.2=log 2 5. Conclusion: this equality is true.

Solve the equation:

log 4 x 4 +log 16 81=log 2 (5x+2). Let's reduce logarithms from the left to the base 2 .

log 2 x 2 +log 2 3=log 2 (5x+2). Take the square root of the number and the base of the first logarithm. Extract the fourth root of the number and the base of the second logarithm.

log 2 (3x 2)=log 2 (5x+2). Convert the sum of logarithms into the logarithm of the product.

3x 2 =5x+2. Received after potentiation.

3x 2 -5x-2=0. We solve a quadratic equation using the general formula for a complete quadratic equation:

a=3, b=-5, c=-2.

D=b 2 -4ac=(-5) 2 -4∙3∙(-2)=25+24=49=7 2 >0; 2 real roots.

Examination.

x=2.

log 4 2 4 +log 16 81=log 2 (5∙2+2);

log 2 2 2 +log 2 3=log 2 12;

log 2 (4∙3)=log 2 12;

log 2 12=log 2 12;

log a n b=(1/ n)∙ log a b

Logarithm of a number b based on a n equal to the product of the fraction 1/ n to the logarithm of a number b based on a.

Find:1) 21log 8 3+40log 25 2; 2) 30log 32 3∙log 125 2 , if it is known that log 2 3=b,log 5 2=c.

Solution.

Solve equations:

1) log 2 x+log 4 x+log 16 x=5.25.

Solution.

Let's reduce these logarithms to base 2. Apply the formula: log a n b=(1/ n)∙ log a b

log 2 x+(½) log 2 x+(¼) log 2 x=5.25;

log 2 x+0.5log 2 x+0.25log 2 x=5.25. Here are similar terms:

(1+0.5+0.25) log 2 x=5.25;

1.75 log 2 x=5.25 |:1.75

log 2 x=3. By definition of logarithm:

2) 0.5log 4 (x-2)+log 16 (x-3)=0.25.

Solution. Let's convert the logarithm to base 16 to base 4.

0.5log 4 (x-2)+0.5log 4 (x-3)=0.25 |:0.5

log 4 (x-2)+log 4 (x-3)=0.5. Let's convert the sum of logarithms into the logarithm of the product.

log 4 ((x-2)(x-3))=0.5;

log 4 (x 2 -2x-3x+6)=0.5;

log 4 (x 2 -5x+6)=0.5. By definition of logarithm:

x 2 -5x+4=0. According to Vieta's theorem:

x 1 =1; x 2 =4. The first value of x will not work, since at x = 1 the logarithms of this equality do not exist, because Only positive numbers can be under the logarithm sign.

Let's check this equation at x=4.

Examination.

0.5log 4 (4-2)+log 16 (4-3)=0.25

0.5log 4 2+log 16 1=0.25

0,5∙0,5+0=0,25

log a b=log c b/log c a

Logarithm of a number b based on A equal to the logarithm of the number b on a new basis With, divided by the logarithm of the old base A on a new basis With.

Examples:

1) log 2 3=lg3/lg2;

2) log 8 7=ln7/ln8.

Calculate:

1) log 5 7, if it is known that lg7≈0,8451; lg5≈0,6990.

c b / log c a.

log 5 7=log7/log5≈0.8451:0.6990≈1.2090.

Answer: log 5 7≈1,209 0≈1,209 .

2) log 5 7 , if it is known that ln7≈1,9459; ln5≈1,6094.

Solution. Apply the formula: log a b =log c b / log c a.

log 5 7=ln7/ln5≈1.9459:1.6094≈1.2091.

Answer: log 5 7≈1,209 1≈1,209 .

Find x:

1) log 3 x=log 3 4+log 5 6/log 5 3+log 7 8/log 7 3.

We use the formula: log c b / log c a = log a b . We get:

log 3 x=log 3 4+log 3 6+log 3 8;

log 3 x=log 3 (4∙6∙8);

log 3 x=log 3 192;

x=192 .

2) log 7 x=lg143-log 6 11/log 6 10-log 5 13/log 5 10.

We use the formula: log c b / log c a = log a b . We get:

log 7 x=lg143-lg11-lg13;

log 7 x=lg143- (lg11+lg13);

log 7 x=lg143-lg (11∙13);

log 7 x=lg143-lg143;

x=1.

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\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)

Let's explain it more simply. For example, \(\log_(2)(8)\) is equal to the power to which \(2\) must be raised to get \(8\). From this it is clear that \(\log_(2)(8)=3\).

Examples:	\(\log_(5)(25)=2\)	because \(5^(2)=25\)
	\(\log_(3)(81)=4\)	because \(3^(4)=81\)
	\(\log_(2)\)\(\frac(1)(32)\) \(=-5\)	because \(2^(-5)=\)\(\frac(1)(32)\)

Argument and base of logarithm

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.”

How to calculate logarithm?

To calculate the logarithm, you need to answer the question: to what power should the base be raised to get the argument?

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 the 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 :

\(\log_(4\sqrt(2))(8)=x\)		We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of a logarithm: \(\log_(a)(c)=b\) \(\Leftrightarrow\) \(a^(b)=c\)
\((4\sqrt(2))^(x)=8\)		What connects \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: \(4=2^(2)\) \(\sqrt(2)=2^(\frac(1)(2))\) \(8=2^(3)\)
\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\)		On the left we use the properties of the degree: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\)
\(2^(\frac(5)(2)x)=2^(3)\)		The bases are equal, we move on to equality of indicators
\(\frac(5x)(2)\) \(=3\)		Multiply both sides of the equation by \(\frac(2)(5)\)
		The resulting root is the value of the logarithm

Answer : \(\log_(4\sqrt(2))(8)=1,2\)

Why was the logarithm invented?

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 as a decimal, it would look like this: \(1.892789260714.....\)

Example : Solve the equation \(4^(5x-4)=10\)

Solution :

\(4^(5x-4)=10\)		\(4^(5x-4)\) 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: \(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
\(\log_(4)(10)=5x-4\)		Let's flip the equation so that X is on the left
\(5x-4=\log_(4)(10)\)		Before us. Let's move \(4\) to the right. And don't be afraid of the logarithm, treat it like an ordinary number.
\(5x=\log_(4)(10)+4\)		Divide the equation by 5
\(x=\)\(\frac(\log_(4)(10)+4)(5)\)		This is our root. Yes, it looks unusual, but they don’t choose the answer.

Answer : \(\frac(\log_(4)(10)+4)(5)\)

Decimal and natural logarithms

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:

Natural logarithm: a logarithm whose base is Euler's number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).

That is, \(\ln(a)\) is the same as \(\log_(e)(a)\)

Decimal Logarithm: A logarithm whose base is 10 is written \(\lg(a)\).

That is, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.

Basic logarithmic identity

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 : \(25\)

How to write a number as a logarithm?

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:

\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)

Any number \(a\) can be represented as a logarithm with base \(b\): \(a=\log_(b)(b^(a))\)

Example : Find the meaning of the expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Solution :

Answer : \(1\)