How to find the value of a logarithmic expression. Logarithmic equation: basic formulas and techniques

21.10.2019

Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information at any time when you contact us.

The following are some examples of the types of personal information we may collect and how we may use such information.

What personal information we collect:

When you submit an application on the site, we may collect various information, including your name, phone number, address Email etc.

How we use your personal information:

The personal information we collect allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
From time to time, we may use your personal information to send you important notices and messages.
We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.

Disclosure to third parties

We do not disclose information received from you to third parties.

Exceptions:

In the event that it is necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from state bodies in the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest purposes.
In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.

Maintaining your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.

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 not defined. Also, 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 base -2 logarithm of 4 is 2.

Basic logarithmic identity

a log a b = b (a > 0, a ≠ 1) (2)

It is important that the domains of definition of the right and left parts of this formula are 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" in solving equations and inequalities can lead to a change in the DPV.

Two obvious consequences of the definition of the logarithm

log a a = 1 (a > 0, a ≠ 1) (3)
log a 1 = 0 (a > 0, a ≠ 1) (4)

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.

The logarithm of the product and the logarithm of the quotient

log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)

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 the thoughtless use of these formulas when solving logarithmic equations and inequalities. When they are used "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 restrict ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of admissible values, and this is categorically unacceptable, since it can lead to the loss of solutions. A similar problem exists for formula (6).

The degree can be taken out of the sign of the logarithm

log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)

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! Taking the power out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of admissible values. All these remarks apply not only to the power of 2, but also to any even power.

Formula for moving to a new base

log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)

That rare case when the ODZ does not change during the conversion. If you have chosen the base c wisely (positive and not equal to 1), the formula for moving to a new base is perfectly safe.

If we choose the number b as a new base c, we obtain an important particular case of formula (8):

Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)

Some simple examples with logarithms

Example 1 Calculate: lg2 + lg50.
Decision. lg2 + lg50 = lg100 = 2. We used the formula for the sum of logarithms (5) and the definition of the decimal logarithm.

Example 2 Calculate: lg125/lg5.
Decision. lg125/lg5 = log 5 125 = 3. We used the new base transition formula (8).

Table of formulas related to logarithms

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)

Logarithm of b (b > 0) to base a (a > 0, a ≠ 1) is the exponent to which you need to raise the number a to get b.

The base 10 logarithm of b can be written as log(b), and the logarithm to the base e (natural logarithm) - ln(b).

Often used when solving problems with logarithms:

Properties of logarithms

There are four main properties of logarithms.

Let a > 0, a ≠ 1, x > 0 and y > 0.

Property 1. Logarithm of the product

Logarithm of the product is equal to the sum of logarithms:

log a (x ⋅ y) = log a x + log a y

Property 2. Logarithm of the quotient

Logarithm of the quotient is equal to the difference of logarithms:

log a (x / y) = log a x – log a y

Property 3. Logarithm of the degree

Degree logarithm is equal to the product of the degree and the logarithm:

If the base of the logarithm is in the exponent, then another formula applies:

Property 4. Logarithm of the root

This property can be obtained from the property of the logarithm of the degree, since the root of the nth degree is equal to the power of 1/n:

The formula for going from a logarithm in one base to a logarithm in another base

This formula is also often used when solving various tasks for logarithms:

Special case:

Comparison of logarithms (inequalities)

Suppose we have 2 functions f(x) and g(x) under logarithms with the same bases and there is an inequality sign between them:

To compare them, you first need to look at the base of the logarithms a:

If a > 0, then f(x) > g(x) > 0
If 0< a < 1, то 0 < f(x) < g(x)

How to solve problems with logarithms: examples

Tasks with logarithms included in the USE in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the relevant sections. Also, tasks with logarithms are found in the bank of tasks in mathematics. You can find all examples by searching the site.

What is a logarithm

Logarithms have always been considered a difficult topic in the school mathematics course. There are many different definitions of the logarithm, but for some reason most textbooks use the most complex and unfortunate of them.

We will define the logarithm simply and clearly. Let's create a table for this:

So, we have powers of two.

Logarithms - properties, formulas, how to solve

If you take the number from the bottom line, then you can easily find the power to which you have to raise a 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 - in fact, the definition of the logarithm:

base a of the argument x is the power to which the number a must be raised to get the number x.

Notation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is equal to.

For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). Might as well 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 row 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 considered 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 segment. 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 indefinitely, and they never repeat. If the logarithm turns out to be irrational, it is better to leave it like this: log 2 5, log 3 8, log 5 100.

It is important to understand that the logarithm is an expression with two variables (base and argument). At first, many people confuse where the base is and where the argument is. To avoid annoying misunderstandings, just take a look at the picture:

Before us is nothing more than the definition of the logarithm. Remember: the logarithm is the power, to which you need to raise the base to get the argument. It is the base that is raised to a power - in the picture it is highlighted in red. It turns out that the base is always at the bottom! I tell this wonderful rule to my students at the very first lesson - and there is no confusion.

How to count logarithms

We figured out the definition - it remains 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:

The argument and base must always be greater than zero. This follows from the definition of the degree by a rational exponent, to which the definition of the logarithm is reduced.
The base must be different from unity, since a unit to any power is still a unit. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!

Such restrictions are called valid range(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) is not imposed. 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 ODZ of the logarithm. All restrictions have already been taken into account by the compilers of the problems. But when logarithmic equations and inequalities come into play, the DHS requirements will become mandatory. Indeed, in the basis and argument there can be very strong constructions that do not necessarily correspond to the above restrictions.

Now consider general scheme logarithm calculations. It consists of three steps:

Express the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
Solve the equation for the variable b: x = a b ;
The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already at the first step. The requirement that the base be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similar to decimals: if you immediately translate them into ordinary ones, there will be many times less errors.

Let's see how this scheme works with specific examples:

Task. Calculate the logarithm: log 5 25

Let's represent the base and the argument as a power of five: 5 = 5 1 ; 25 = 52;

Let's make and solve the equation:
log 5 25 = b ⇒(5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2;

Received an answer: 2.

Task. Calculate the logarithm:

Task. Calculate the logarithm: log 4 64

Let's represent the base and the argument as a power of two: 4 = 2 2 ; 64 = 26;
Let's make and solve the equation:
log 4 64 = b ⇒(2 2) b = 2 6 ⇒2 2b = 2 6 ⇒2b = 6 ⇒ b = 3;
Received an answer: 3.

Task. Calculate the logarithm: log 16 1

Let's represent the base and the argument as a power of two: 16 = 2 4 ; 1 = 20;
Let's make and solve the equation:
log 16 1 = b ⇒(2 4) b = 2 0 ⇒2 4b = 2 0 ⇒4b = 0 ⇒ b = 0;
Received a response: 0.

Task. Calculate the logarithm: log 7 14

Let's represent the base and the argument as a power of seven: 7 = 7 1 ; 14 is not represented as a power of seven, because 7 1< 14 < 7 2 ;
It follows from the previous paragraph that the logarithm is not considered;
The answer is no change: log 7 14.

A small note on the last example. How to make sure that a number is not an exact power of another number? Very simple - just decompose it into prime factors. If there are at least two distinct factors in the expansion, the number is not an exact power.

Task. Find out if the exact powers of the number are: 8; 48; 81; 35; fourteen.

8 \u003d 2 2 2 \u003d 2 3 - the exact degree, because there is only one multiplier;
48 = 6 8 = 3 2 2 2 2 = 3 2 4 is not an exact power because there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 \u003d 3 4 - exact degree;
35 = 7 5 - again not an exact degree;
14 \u003d 7 2 - again not an exact degree;

We also note that we prime numbers are always exact powers of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and designation.

of the x argument is the base 10 logarithm, i.e. the power to which 10 must be raised to obtain x. Designation: lgx.

For example, log 10 = 1; log 100 = 2; lg 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” appears in the textbook, know that this is not a typo. This is the decimal logarithm. However, if you are not used to such a designation, you can always rewrite it:
log x = log 10 x

Everything that is true for ordinary logarithms is also true for decimals.

natural logarithm

There is another logarithm that has its own notation. In a sense, it is even more important than decimal. This is the natural logarithm.

of the x argument is the logarithm to the base e, i.e. the power to which the number e must be raised to get the number x. Designation: lnx.

Many will ask: what is the number e? This is an irrational number, its exact value cannot be found and written down. Here are just the first numbers:
e = 2.718281828459…

We will not delve into 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; log 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 irrational. Except, of course, unity: ln 1 = 0.

For natural logarithms, all the rules that are true for ordinary logarithms are valid.

Logarithm. Properties of the logarithm (power of the logarithm).

How to represent a number as a logarithm?

We use the definition of a logarithm.

The logarithm is an indicator of the power to which the base must be raised to get the number under the sign of the logarithm.

Thus, in order to represent a certain number c as a logarithm to the base a, it is necessary to put a degree under the sign of the logarithm with the same base as the base of the logarithm, and write this number c into the exponent:

In the form of a logarithm, you can represent absolutely any number - positive, negative, integer, fractional, rational, irrational:

In order not to confuse a and c in stressful conditions of a test or exam, you can use the following rule to remember:

what is below goes down, what is above goes up.

For example, you want 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, in the base of the degree, and which - up, in the exponent.

The base 3 in the record of the logarithm is at the bottom, which means that when we represent the deuce as a logarithm to the base of 3, we will also write 3 down to the base.

2 is higher than 3. And in the notation of the degree, we write the two above the three, that is, in the exponent:

Logarithms. First level.

Logarithms

logarithm positive number b by reason a, where a > 0, a ≠ 1, is 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. He is usually called logarithmic identity.
The action of finding the logarithm of a number is called logarithm.

Properties of logarithms:

The logarithm of the product:

Logarithm of the quotient from division:

Replacing the base of the logarithm:

Degree logarithm:

root logarithm:

Logarithm with power base:

Decimal and natural logarithms.

Decimal logarithm numbers call the base 10 logarithm of that number and write lg b
natural logarithm numbers call the logarithm of this number to the base e, where e is an irrational number, approximately equal to 2.7. At the same time, they write ln b.

Other Notes on Algebra and Geometry

Basic properties of logarithms

Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same base: log a x and log a y. Then they can be added and subtracted, and:

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

So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Note: key moment here - same grounds. If the bases are different, these rules do not work!

These formulas will help calculate the 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 the bases of logarithms are the same, 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 considered separately. But after transformations quite normal numbers turn out. Based on this fact, many test papers. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? 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 their 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 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 sign of the logarithm into the logarithm itself.

How to solve logarithms

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 according to the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 2 4 ; 49 = 72. We have:

I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have 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 was done. The result is the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?

Formulas for transition to a new base come to the rescue. We 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 put c = x, we get:

It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

Task. Find the value of the expression: log 5 16 log 2 25.

Note that the arguments of both logarithms are exact exponents. 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 flip the second logarithm:

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the 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 it down and get rid of the indicators:

Now let's get rid of the decimal logarithm by moving to a new base:

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base.

In this case, the 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's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:

Note that log 25 64 = log 5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.

log a a = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
log a 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a 0 = 1 is a direct consequence of the definition.

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.

Today we will talk about logarithm formulas and give demonstration solution examples.

By themselves, they imply solution patterns according to the basic properties of logarithms. Before applying the logarithm formulas to the solution, we recall for you, first all the properties:

Now, based on these formulas (properties), we show examples of solving logarithms.

Examples of solving logarithms based on formulas.

Logarithm a positive number b in base a (denoted log a b) is the 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, so 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 is an ordinary logarithm, the base of which is 10. Denoted as lg.

log 10 100 = 2 because 10 2 = 100

natural logarithm- also the usual logarithm logarithm, but with the base e (e \u003d 2.71828 ... - an irrational number). Referred to as ln.

It is desirable to remember 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.

Basic logarithmic identity
a log a b = b
8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
The logarithm of the product is equal to the sum of the logarithms
log a (bc) = log a b + log a c
log 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
The logarithm of the quotient is equal to the difference of the logarithms
log a (b/c) = log a b - log a c
9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
Properties of the degree of a logarithmable number and the base of the logarithm
The exponent of a logarithm 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
Transition to a new foundation
log a b = log c b / log c a,
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 logarithm formulas are not as complicated as they seem. Now, having considered examples of solving logarithms, we can move on to logarithmic equations. We will consider 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: decided to get an education of another class study abroad as an option.

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 indicators. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where it is required to simplify cumbersome multiplication to simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. Simple and accessible language.

Definition in mathematics

The 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" by its base "a" is considered the power of "c", to which the base "a" must be raised, so that in the end 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 such a degree that from 2 to the required degree you get 8. Having done some calculations in your mind, we get the number 3! And rightly so, because 2 to the power of 3 gives the number 8 in the answer.

Varieties 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 distinct kinds of logarithmic expressions:

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

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values of logarithms, one should remember their properties and the order of actions in their decisions.

Rules and some restrictions

In mathematics, there are several rules-limitations that are accepted as an axiom, that is, they are not subject to discussion and are true. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the root of an even degree from negative numbers. Logarithms also have their own rules, following which you can easily learn how to work even with long and capacious logarithmic expressions:

the base "a" must always be greater than zero, and at the same time not be 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 be greater than zero.

How to solve logarithms?

For example, given the task to find the answer to the equation 10 x \u003d 100. It is very easy, you need to choose such a power by raising the number ten to which we get 100. This, of course, is 10 2 \u003d 100.

Now let's represent this expression as a logarithmic one. We get log 10 100 = 2. When solving logarithms, all actions practically converge to finding the degree to which the base of the logarithm must be entered in order to obtain a given number.

To accurately determine the value of an unknown degree, you must 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 mindset and knowledge of the multiplication table. However, for large values you need a table of degrees. It can be used even by those who do not understand anything at all in complex mathematical topics. The left column contains numbers (base a), top row of numbers is the value of the power c to which the number a is raised. At the intersection in the cells, the values of the numbers are determined, which 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 real 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 equation. For example, 3 4 =81 can be written as the logarithm of 81 to base 3, which is four (log 3 81 = 4). For negative powers, the rules are the same: 2 -5 = 1/32 we write 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 consider examples and solutions of equations a little lower, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.

An expression of the following form is given: log 2 (x-1) > 3 - it is a 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 in 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 of 2 x = √9) imply one or more specific numerical values in the answer, while when solving the inequality, both the range of acceptable values and the points breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer of the equation, but a continuous series or set of numbers.

Basic theorems about logarithms

When solving primitive tasks on 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 get acquainted with examples of equations later, let's first analyze each property in more detail.

The basic identity looks like this: a logaB =B. It only applies if 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 prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this formula of logarithms, with examples and a 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 get that s 1 *s 2 = a f1 *a f2 = a f1+f2 (degree properties ), and further by definition: log a (s 1 *s 2)= f 1 + f 2 = log a s1 + log a s 2, which was to be proved.
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 "property of the degree of the logarithm". It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics rests on regular postulates. Let's look at the proof.

Let log a b \u003d t, it turns out a t \u003d b. If you raise both parts to the power m: a tn = b n ;

but since a tn = (a q) nt/q = b n , hence 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 logarithm problems are examples of equations and inequalities. They are found in almost all problem books, and are also included in the mandatory part of exams in mathematics. To enter a university or pass entrance tests in mathematics, you need to know how to solve such tasks correctly.

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, 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 general view. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them soon.

When solving logarithmic equations, it is necessary to determine what type of logarithm we have before us: an example of an expression may contain a natural logarithm or a decimal one.

Here are examples ln100, ln1026. Their solution boils down to the fact that you need to determine the degree to which the base 10 will be equal to 100 and 1026, respectively. For solutions of natural logarithms, one must 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 main theorems on logarithms.

The property of the logarithm of the 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 degree of the logarithm, we managed to solve at first glance a complex and unsolvable expression. It is only necessary to factorize the base and then take the exponent values out of the sign of the logarithm.

Tasks from the exam

Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam (state exam for all school graduates). Usually these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most difficult and voluminous tasks). The exam implies an accurate and perfect knowledge of the topic "Natural logarithms".

Examples and problem solutions are taken from official USE 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.

All logarithms are best reduced to the same base so that the solution is not cumbersome and confusing.
All expressions under the sign of the logarithm are indicated as positive, therefore, when taking out the exponent of the exponent of the expression, which is under the sign of the logarithm and as its base, the expression remaining under the logarithm must be positive.