2. Solve the equation $\log _{a}(1+\sqrt{x})=\log _{b} x$, where $a>1, b>1, a^{2}=b+1$.
Solution. Let $\log _{a}(1+\sqrt{x})=\log _{b} x=t$. We have $1+\sqrt{x}=a^{t}$ and $x=b^{t}$

$\Rightarrow 1+\sqrt{b^{t}}=a^{t} \Leftrightarrow 1+b^{t / 2}=\left(a^{2}\right)^{1 / 2}$. If we let $t / 2=u$, the equation can be written as $1+b^{u}=\left(a^{2}\right)^{u}$, that is, $1+b^{u}=(1+b)^{u} \Leftrightarrow\left(\frac{1}{1+b}\right)^{u}+\left(\frac{b}{1+b}\right)^{u}=1$.

The function $f: \mathbb{R} \rightarrow \mathbb{R}, f(u)=\left(\frac{1}{1+b}\right)^{u}+\left(\frac{b}{1+b}\right)^{u}$ is decreasing and since $f(u)=f(1)=1$, it follows that $u=1 \Rightarrow t=2 \Rightarrow x=b^{2}$, the unique solution.