4.a) Solve the equation: $3^{x}=2^{x}+1$.

b) Solve in $(1, \infty)$ the equation: $x^{\log _{3} 2}+1=(x-1)^{\log _{2} 3}$.

## Note:

All subjects are mandatory;

Working time: 3 hours.

## Solutions for class aX-a:
4.a) $3^{x}=2^{x}+1 \Rightarrow\left(\frac{2}{3}\right)^{x}+\left(\frac{1}{3}\right)^{x}=1$.

We observe that $x=1$ is a solution to the equation. We show that this solution is unique.

The function $g: \mathbb{R} \rightarrow(0, \infty), g(x)=\left(\frac{2}{3}\right)^{x}+\left(\frac{1}{3}\right)^{x}$, is strictly decreasing, hence injective, and thus the unique solution is $x=1$.
b) $x^{\log _{3} 2}=2^{\log _{3} x}$ and $(x-1)^{\log _{2} 3}=3^{\log _{2}(x-1)}$.

Let $a=\log _{3} x$ and $b=\log _{2}(x-1) \Rightarrow x=3^{a}$ and $x=2^{b}+1$,

From here we obtain: $3^{a}=2^{b}+1$.

The given equation is equivalent to: $2^{a}+1=3^{b}$.

From the two relations, we obtain:

$$
3^{a}-2^{b}=3^{b}-2^{a} \Leftrightarrow 3^{a}+2^{a}=3^{b}+2^{b}
$$

Consider the function: $f: \mathbb{R} \rightarrow(0, \infty), f(x)=3^{x}+2^{x}$, which is strictly increasing and thus injective, so $a=b$.

Then: $3^{a}=2^{a}+1 \Rightarrow\left(\frac{2}{3}\right)^{a}+\left(\frac{1}{3}\right)^{a}=1$.

The function $g: \mathbb{R} \rightarrow(0, \infty), g(x)=\left(\frac{2}{3}\right)^{x}+\left(\frac{1}{3}\right)^{x}$, is strictly decreasing, hence injective, and thus the unique solution is $a=1 \Rightarrow x=3$.