## SUBJECT 3

Solve the equation $2 \cdot \log _{3}\left(x+\frac{2}{\sqrt[3]{x}}\right)=\sqrt{x+3}-\sqrt{x-1}$.

Prof. Niculae Cavachi
## SUBJECT 3

The domain is $[1 ; \infty)$. We have $\sqrt{x+3}-\sqrt{x-1}=\frac{4}{\sqrt{x+3}+\sqrt{x-1}} \leq 2$, since $\sqrt{x+3}+\sqrt{x-1} \geq 2, \forall x \geq 1$ as a strictly increasing function. $3 p$ $\log _{3}\left(x+\frac{2}{\sqrt[3]{x}}\right) \leq 1 \Leftrightarrow x+\frac{2}{\sqrt[3]{x}} \leq 3$ $1 p$

Let $\sqrt[3]{x}=u, u \geq 1 \Rightarrow u^{4}-3 u+2 \leq 0 \Rightarrow(u-1)\left(u^{3}+u^{2}+u-2\right) \leq 0$ $.2 p$

but $u^{3}+u^{2}+u-2 \geq 1$ and thus $(u-1) \leq 0 \Leftrightarrow u \leq 1 \Rightarrow u=1 \Rightarrow x=1$ unique solution $.1 \mathrm{p}$