B3. Solve the exponential equation $\frac{2^{x}+2^{-x}}{2}=2\left(2^{x}-2^{-x}\right)$. The solution should be exact.

19th Knowledge Competition

in mathematics for students of secondary technical and vocational schools

National Competition, April 13, 2019
Attach the sticker with the code

## Tasks for 4th year

Time for solving: 120 minutes. In section A, we will award three points for each correct answer, and deduct one point for each incorrect answer. Enter the answers for section A in the left table, leave the right table blank.

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-14.jpg?height=163&width=319&top_left_y=661&top_left_x=640)

| B1 | B2 | B3 |
| :--- | :--- | :--- |
|  |  |  |
B3. In the exponential equation, we first eliminate the fraction, so we multiply by 2 and get $2^{x}+2^{-x}=4\left(2^{x}-2^{-x}\right)$, remove the parentheses: $2^{x}+2^{-x}=4 \cdot 2^{x}-4 \cdot 2^{-x}$, move all terms to one side: $5 \cdot 2^{-x}-3 \cdot 2^{x}=0$ and note that $2^{-x}=\frac{1}{2^{x}}$. We rewrite the equation as $\frac{5}{2^{x}}-3 \cdot 2^{x}=0$, introduce a new variable $2^{x}=t$ and get the equation $\frac{5}{t}-3 \cdot 5=0$; multiply by $t$ and write the solution $5-3 t^{2}=0, t= \pm \frac{\sqrt{15}}{3}$. Since $t=2^{x}$, we consider only the positive solution $2^{x}=\frac{\text { sqrt15 }}{3}$. We logarithmize the exponential equation: $\log _{2} 2^{x}=\log _{2}\left(\frac{\sqrt{15}}{3}\right)$ and write the solution $x=\log _{2}\left(\frac{\sqrt{15}}{3}\right)^{3}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=65&width=1731&top_left_y=573&top_left_x=174)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=54&width=1739&top_left_y=624&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=63&width=1742&top_left_y=677&top_left_x=174)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=54&width=1742&top_left_y=727&top_left_x=171)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=66&width=1742&top_left_y=778&top_left_x=174)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=66&width=1739&top_left_y=841&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-25.jpg?height=89&width=1739&top_left_y=892&top_left_x=176)

The solution to the exponential equation $x=\log _{2}\left(\frac{\sqrt{15}}{3}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

## 19th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools  National Competition, April 13, 2019

## Solutions to Problems and Scoring

(5. APRIL 2019, $10: 30$ )

A competitor who arrives at the solution by any correct method (even if the scoring guide does not anticipate it) receives all possible points.

A correct method is considered any procedure that:

- sensibly takes into account the problem statement,
- leads to the solution of the problem,
- is mathematically correct and complete.

If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided for. If, however, the solution is guessed (cannot be reached without calculation), even if it is correct, it is scored with 0 points.

A competitor who has only partially solved the problem, from a correct but incomplete solution process, cannot receive more than half of the possible points.

The asterisk '*' next to the points means that the competitor can receive the point or points for a correct procedure, even if the calculation is incorrect.

## Fourth Year

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-27.jpg?height=140&width=223&top_left_y=358&top_left_x=931)