## Problem 1

Existence conditions $\mathrm{x} \in(27,+\infty)$

$\log _{\frac{1}{3}}(\sqrt[3]{x}-3)=\mathrm{a} \Rightarrow\left(\frac{1}{3}\right)^{a}=\sqrt[3]{x}-3$

and $\lg (\sqrt[3]{x}+4)=b \Rightarrow 10^{b}=\sqrt[3]{x}+4$

From (1) and (2) $\Rightarrow\left(\frac{1}{3}\right)^{a}-10^{b}=-7$

(1 p)

With the notations made, the equation from the statement becomes: $\left(\frac{1}{10}\right)^{a}-3^{b}=7$

Adding relations (3) and (4) we get: $\left(\frac{1}{3}\right)^{a}-3^{b}+\left(\frac{1}{10}\right)^{a}-10^{b}=0 \Leftrightarrow\left(3^{-a}-3^{b}\right)+\left(10^{-a}-10^{b}\right)=0$.

(1 p)

If $-\mathrm{ab}, 3^{-\mathrm{a}}>3^{\mathrm{b}}, 10^{-\mathrm{a}}>10^{\mathrm{b}}$ and $\left(3^{-\mathrm{a}}-3^{\mathrm{b}}\right)+\left(10^{-\mathrm{a}}-10^{\mathrm{b}}\right)>0 \quad$ (1 p)

Therefore $-\mathrm{a}=\mathrm{b}$. Equation (4) becomes: $\left(\frac{1}{10}\right)^{-\mathrm{b}}-3^{\mathrm{b}}=7$ or $10^{\mathrm{b}}-3^{\mathrm{b}}=7$.

It is observed that $\mathrm{b}=1$ is a solution.

(1 p)

$10^{\mathrm{b}}-3^{\mathrm{b}}=7 \Rightarrow 7+3^{b}=10^{b} \Rightarrow 7 \cdot\left(\frac{1}{10}\right)^{b}+\left(\frac{3}{10}\right)^{b}=1$

The function $\mathrm{f}:(\lg 7,+\infty) \rightarrow(0,+\infty), f(b)=7 \cdot\left(\frac{1}{10}\right)^{b}+\left(\frac{3}{10}\right)^{b}$ is strictly decreasing, hence injective, and the equation $f(b)=1$ will have the unique solution $b=1$.

For $\mathrm{b}=1$ we have $\sqrt[3]{\mathrm{x}}=10^{\mathrm{b}}-4=10-4=6, x=6^{3}=216 \in(27,+\infty) \quad$ (1 p)
## Problem 1

Solve the equation $\left(\frac{1}{10}\right)^{\log _{\frac{1}{3}}(\sqrt[3]{x}-3)}-3^{\lg (\sqrt[3]{x}+4)}=7$.