B4. Calculate the integer solution of the equation $\log _{2}\left(x^{2}+7\right)=\frac{\log _{2}(x+1)}{\log _{8} 2}-\log _{2} x$. (7 points)

Space for solving problems in set $B$.

## dIFA

12th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools

Regional Competition, March 28, 2012

## PROBLEMS FOR THE FOURTH YEAR

Time for solving: 90 minutes. In set A, we will evaluate a correct answer with two points, while we will deduct half a point for an incorrect answer. Write the answers for set A in the left table.
![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-08.jpg?height=172&width=1210&top_left_y=658&top_left_x=433)
B4. We use the rules for calculating with logarithms: $\log _{8} 2=\frac{1}{3}$, so the right side of the equation is equal to $3 \log _{2}(x+1)-\log _{2} x$. On the right side, we also use the rule for the logarithm of a power, so the right side of the equation is equal to: $\log _{2}(x+1)^{3}-\log _{2} x$. We transform the difference into $\log _{2} \frac{(x+1)^{3}}{x}$. When we antilogarithmize the equation, we get $x^{2}+7=\frac{(x+1)^{3}}{x}$. We rearrange the equation and get the quadratic equation $3 x^{2}-4 x+1=0$. The solutions to the equation are $x_{1}=1$ and $x_{2}=\frac{1}{3}$. The integer solution to the equation is 1.

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-16.jpg?height=78&width=1687&top_left_y=2194&top_left_x=230)

The simplified right side of the equation $\log _{2}(x+1)^{3}-\log _{2} x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-16.jpg?height=57&width=1685&top_left_y=2307&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-16.jpg?height=60&width=1685&top_left_y=2357&top_left_x=228)

Solving the equation ..................................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-16.jpg?height=66&width=1687&top_left_y=2457&top_left_x=230)

Answer: The integer solution to the equation is $1 \ldots \ldots \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$. point

## Fourth Year

## Group A

| A1 | A2 | A3 | A4 | A5 | A6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| C | C | D | D | E | E |