## Task B-3.7.

A square with an area of $P=7-\log _{2} x$ and a cube with a volume of $V=\log _{2} x-2$ are given. Calculate the real number $x$ if the side length of the square is 1 unit greater than the edge length of the cube. What are the side length of the square and the edge length of the cube?
## Solution.

Let $a$ be the length of the edge of the cube, and $b$ the length of the side of the square. Then

$V=\log _{2} x-2=a^{3}$, so $a=\sqrt[3]{\log _{2} x-2}$.

$P=7-\log _{2} x=b^{2}$, so $b=\sqrt{7-\log _{2} x}$.

Since $a$ and $b$ are lengths, it must hold that $\log _{2} x-2>0$ and $7-\log _{2} x>0$, so $2<\log _{2} x<7$ (or $4<x<128$).

According to the condition of the problem, we need to solve the equation

$$
\sqrt{7-\log _{2} x}-\sqrt[3]{\log _{2} x-2}=1
$$

If we introduce the substitution $t=\log _{2} x$, we get a series of equivalent equations:

$$
\begin{array}{cc}
\sqrt{7-t}-\sqrt[3]{t-2}=1 & \\
\sqrt{7-t}-1=\sqrt[3]{t-2} & \\
(\sqrt{7-t}-1)^{3}=t-2 & 1 \text { point } \\
(7-t) \sqrt{7-t}-3(7-t)+3 \sqrt{7-t}-1=t-2 & \\
(10-t) \sqrt{7-t}=20-2 t & 1 \text { point } \\
((10-t) \sqrt{7-t})^{2}=(20-2 t)^{2} & 1 \text { point } \\
(10-t)^{2}(7-t)=4(10-t)^{2} & 1 \text { point } \\
(10-t)^{2}(3-t)=0 &
\end{array}
$$

The solutions to the last equation are $t=10$ and $t=3$,

so $\log _{2} x=10$ or $\log _{2} x=3$.

Due to the condition, the first equation does not lead to a solution of our problem. 1 point

$\begin{array}{ll}\text { From } \log _{2} x=3 \text { it follows that } x=8 . & 1 \text { point }\end{array}$

The sought lengths of the sides of the square and the edge of the cube are $b=2$ and $a=1 . \quad 1$ point

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
January 5, 2019.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.