home

How to solve the problem of the exam number 13 on exponential and logarithmic equations | 1C: Tutor

What you need to know about exponential and logarithmic equations for solving problems of the exam in mathematics?

Being able to solve exponential and logarithmic equations is very important for successfully passing a single state examination in mathematics of the profile level. Important for two reasons:

First of all, task No. 13 of the KIM version of the USE, albeit infrequently, but still sometimes represents just such an equation that needs not only to be solved, but also (similar to the task in trigonometry) to select the roots of the equation that satisfy some condition.

So, one of the options for 2017 included the following task:

a) Solve the equation 8 x – 7 . 4 x – 2 x +4 + 112 = 0.

b) Indicate the roots of this equation that belong to the segment.

Answer: a) 2; log 2 7 and b) log 2 7.

In another version, there was such a task:

a) Solve the equation 6log 8 2 x - 5log 8 x + 1 = 0

b) Find all the roots of this equation that belong to the segment.

Answer: a) 2 and 2√ 2 ; b) 2.

There was also this:

a) Solve the equation 2log 3 2 (2cos x) - 5log 3 (2cos x) + 2 = 0.

b) Find all roots of this equation that belong to the segment [π; 5π / 2].

Answer: and) (π / 6 + 2πk; -π / 6 + 2πk, k∊Z) and b) 11π / 6; 13π / 6.

Secondly, the study of methods for solving exponential and logarithmic equations is good, since the basic methods for solving both equations and inequalities actually use the same mathematical ideas.

The main methods for solving exponential and logarithmic equations are easy to remember, there are only five of them: reduction to the simplest equation, the use of equivalent transitions, the introduction of new unknowns, logarithm and factorization. The method of using the properties of exponential, logarithmic and other functions in solving problems is worth a separate one: sometimes the key to solving an equation is the domain of definition, the range of values, non-negativity, boundedness, parity of the functions included in it.

As a rule, in Problem No. 13 there are equations that require the use of the above five basic methods. Each of these methods has its own characteristics that you need to know, since it is their ignorance that leads to errors in solving problems.

What are the typical mistakes test takers make?

Often, when solving equations containing an exponential function, schoolchildren forget to consider one of the cases of equality. As is known, equations of this type are equivalent to a set of two systems of conditions (see below), we are talking about the case when a ( x) = 1

This error is due to the fact that solving the equation the examinee formally uses the definition of the exponential function (y \u003d ax, a\u003e 0, a ≠ 1): for and ≤ 0 the exponential function is not really defined,

But with and = 1 is defined, but not indicative, since the unit in any real degree is identically equal to itself. This means that if in the considered equation at and(x) = 1 there is a correct numerical equality, then the corresponding values \u200b\u200bof the variable will be the roots of the equation.

Another mistake is applying the properties of logarithms without taking into account the range of acceptable values. For example, the well-known property "the logarithm of the product is equal to the sum of the logarithms", it turns out, has a generalization:
log a ( f(x)g(x)) \u003d log a │ f(x) │ + log a │g ( x) │, for f(x)g(x) > 0, a > 0, a ≠ 1

Indeed, for the expression on the left-hand side of this equality to be defined, it is sufficient that the product of functions f and g was positive, but the functions themselves can be both simultaneously greater and simultaneously less than zero, therefore, when applying this property, it is necessary to use the concept of a module.

And there are many such examples. Therefore, to effectively master the methods of solving exponential and logarithmic equations, it is best to use the services that will be able to tell about such "pitfalls" by examples of solving the corresponding examination problems.

Practice problem solving regularly

To start practicing on the 1C: Tutor portal, it is enough.
You can:

All courses consist of a methodologically correct sequence of theory and practice necessary for successful problem solving. Includes theory in the form of texts, slides and videos, problems with solutions, interactive simulators, models, and tests.

Still have questions? Call us at 8 800 551-50-78 or write to online chat.

Here are the key phrases to help search engines find our tips:
How to solve task 13 in exam exam, problems for logarithms, Kim USE 2017, preparation for the USE profile mathematician, Mathematics profile, solving equations and logarithms, solving problems for exponential equations of the USE, calculating the properties of logarithms, exponential function, problems in mathematics of the profile level, applying the properties of logarithms, solving problems at the roots, the tasks of the exam 2017 on exponential equations, preparation for exam graduates Grade 11 in 2018, entering a technical university.

Back forward

Attention! The slide preview is used for informational purposes only and may not represent all the presentation options. If you are interested in this work, please download the full version.

A method of solution is good if from the very beginning we can foresee - and subsequently confirm it -
that by following this method we will achieve the goal.
G. Leibniz

LESSON TYPE: Consolidation and improvement of knowledge.

• Didactic - Repeat and consolidate the properties of logarithms; logarithmic equations; fix the methods for solving the largest and smallest function values; improve the application of the knowledge gained in solving the problems of the exam C1 and C3;
• Developing - Development of logical thinking, memory, cognitive interest, continue the formation of mathematical speech and graphic culture, develop the ability to analyze;
• Educational - To teach the aesthetic design of writing in a notebook, the ability to communicate, instill accuracy.

Equipment: blackboard, computer, projector, screen, cards with test tasks, with tasks for the work of all students.

Forms of work: fhorizontal, individual, collective.

DURING THE CLASSES

1. ORGANIZATIONAL MOMENT

2. PURPOSE STATEMENT

3. CHECKING THE HOMEWORK

4. KNOWLEDGE UPDATE

Analyze: in which tasks of the exam there are logarithms.

(B-7-simplest logarithmic equations

B-11-transformation of logarithmic expressions

B-12- problems of physical content related to logarithms

B-15- finding the largest and smallest function value

С-1- trigonometric equations containing logarithm

С-3 - a system of inequalities containing a logarithmic inequality)

At this stage, oral work is carried out, during which students not only remember the properties of logarithms, but also perform the simplest USE tasks.

1) Definition of the logarithm. What properties of the logarithm do you know? (and conditions?)

1.log b b \u003d 1
2.log b 1 \u003d 0, 3.log c (ab) \u003d log c a + log c b.
4.log c (a: b) \u003d log c a - log c b.
5.log c (b k) \u003d k * log c

2) What function is called logarithmic? D (y) -?

3) What is the decimal logarithm? ()

4) What is natural logarithm? ()

5) What is the number e?

6) What is the derivative of? ()

7) What is the derivative of the natural logarithm?

5. ORAL WORK for all students

Calculate orally: (tasks B-11)

= = = =

152 1 144 -1/2

6. Independent activity of students in solving tasks

B-7 followed by verification

Solve the equations (the first two equations are spoken orally, and the rest is solved independently by the whole class and writes down the solution in a notebook):

(While the students work on the spot on their own, 3 students go to the board and work on individual cards)

After checking 3-5 equations from the spot, the children are invited to prove that the equation has no solution (orally)

7. Solution B-12 - (physical content problems related to logarithms)

The whole class solves the problem (there are 2 people at the board: the 1st one solves together with the class, the 2nd one solves a similar problem independently)

8. ORAL WORK (questions)

Recall the algorithm for finding the largest and smallest values \u200b\u200bof a function on a segment and on an interval.

Work on the board and in a notebook.

(prototype B15 - Unified State Exam)

9. Mini-test with self-control.

№	Option 1	Option 2
1.	=
2.
3.
4.
5.
6.	Find the largest function value 11. Students act as experts The children are invited to evaluate the student's work - task S-1, completed on the examination form - 0.1.2 points (see presentation) 12. HOME TASK The teacher explains homework, paying attention to the fact that similar tasks were considered in the lesson. Students listening carefully to the teacher's explanations, write down their homework. FIPI ( open bank assignments: section geometry, 6th page) uztest.ru (logarithm transformation) С3 - the task of the second part of the exam 13. SUMMING UP Today in the lesson we have repeated the properties of logarithms; logarithmic equations; fixed methods for finding the highest and lowest value of a function; examined the problems of physical content related to logarithms; solved problems C1 and C3, which are offered at the exam in mathematics in prototypes B7, B11, B12, B15, C1 and C3. Grading.

In task number 12 of the USE in mathematics of the profile level, we need to find the largest or smallest value of the function. For this, it is necessary to use, obviously, a derivative. Let's look at a typical example.

Analysis of typical options for assignments No. 12 of the USE in mathematics of the profile level

The first variant of the task (demo version 2018)

Find the maximum point of the function y \u003d ln (x + 4) 2 + 2x + 7.

Solution algorithm:

1. Find the derivative.
2. We write down the answer.

Decision:

1. We are looking for values \u200b\u200bof x for which the logarithm makes sense. To do this, we solve the inequality:

Since the square of any number is non-negative. The solution to the inequality will be only that value of x for which x + 4 ≠ 0, i.e. for x ≠ -4.

2. Find the derivative:

y '\u003d (ln (x + 4) 2 + 2x + 7)'

By the property of the logarithm, we get:

y '\u003d (ln (x + 4) 2)' + (2x) '+ (7)'.

By the formula for the derivative of a complex function:

(lnf) ’\u003d (1 / f) ∙ f’. We have f \u003d (x + 4) 2

y, \u003d (ln (x + 4) 2) '+ 2 + 0 \u003d (1 / (x + 4) 2) ∙ ((x + 4) 2)' + 2 \u003d (1 / (x + 4) 2 2) ∙ (x 2 + 8x + 16) '+ 2 \u003d 2 (x + 4) / ((x + 4) 2) + 2

y '\u003d 2 / (x + 4) + 2

3. Equate the derivative to zero:

y, \u003d 0 → (2 + 2 ∙ (x + 4)) / (x + 4) \u003d 0,

2 + 2x +8 \u003d 0, 2x + 10 \u003d 0,

Second variant of the task (from Yashchenko, no. 1)

Find the minimum point of the function y \u003d x - ln (x + 6) + 3.

Solution algorithm:

1. We define the scope of the function.
2. Find the derivative.
3. Determine at what points the derivative is 0.
4. We exclude points that do not belong to the definition area.
5. Among the remaining points, we are looking for the values \u200b\u200bof x at which the function has a minimum.
6. We write down the answer.

Decision:

1. ODZ:.

2. Find the derivative of the function:

3. Equate the resulting expression to zero:

4. Received one point x \u003d -5, belonging to the domain of the function.

5. At this point, the function has an extremum. Let's check if this is the minimum. When x \u003d -4

When x \u003d -5.5, the derivative of the function is negative, since

Hence, the point x \u003d -5 is the minimum point.

The third variant of the task (from Yashchenko, no. 12)

Solution algorithm:

1. Find the derivative.
2. Determine at what points the derivative is 0.
3. We exclude points that do not belong to the specified segment.
4. Among the remaining points, we are looking for the values \u200b\u200bof x at which the function has a maximum.
5. Find the values \u200b\u200bof the function at the ends of the segment.
6. We are looking for the largest among the obtained values.
7. We write down the answer.

Decision:

1. We calculate the derivative of the function, we get