## Subject II. (30 points)

Solve the equation: $13^{x^{2}} \cdot 25^{x^{2}+2 x+\frac{3}{2}} \cdot 31^{x^{2}+x+1}=2015$.

Prof. Raul Domsa, "Petru Maior" Theoretical High School, Gherla
## Subject II. (30 points)

$$
\begin{aligned}
& 13^{x^{2}} \cdot 25^{x^{2}+2 x+\frac{3}{2}} \cdot 31^{x^{2}+x+1}=2015 /:(13 \cdot 5 \cdot 31) \Rightarrow \\
& 13^{x^{2}-1} \cdot 25^{x^{2}+2 x+1} \cdot 31^{x^{2}+x}=1 \Leftrightarrow 13^{(x+1)(x-1)} \cdot 25^{(x+1)^{2}} \cdot 31^{x(x+1)}=1 \Leftrightarrow\left[13^{x-1} \cdot 25^{x+1} \cdot 31^{x}\right]^{(x+1)}=1
\end{aligned}
$$

We take the logarithm of both sides: (x+1) $\cdot \lg \left(13^{x-1} \cdot 25^{x+1} \cdot 31^{x}\right)=0 \Rightarrow x_{1}=-1$ and $(x-1) \cdot \lg 13+(x+1) \cdot \lg 25+x \cdot \lg 31=0$

$$
\Rightarrow x_{2}=\frac{\lg 13-\lg 25}{\lg 13+\lg 25+\lg 31}
$$