Solve the logarithmic equation

Let $a=\sqrt{2}-1$ and $b=x+1$ . Then the equation becomes:

$\displaystyle{\frac{\ln{a}}{\ln{b}}\cdot \frac{\ln(b^2+1)}{\ln(a^2+1)}}=1$

or equivalently:

$\displaystyle{\frac{\ln{a}}{\ln(a^2+1)}=\frac{\ln b}{\ln(b^2+1)}}$

Now, the derivative of $\frac{\ln{t}}{\ln(t^2+1)}$ is

$\frac{\frac{\ln(t^2+1)}{t}-\frac{2t\ln t}{t^2+1}}{\ln^2(t^2+1)} = \frac{(t^2+1)\ln(t^2+1)-t^2\ln t^2}{t(t^2+1)\ln^2(t^2+1)} > 0$

So we conclude that the only solution is $a=b$ , or:

$x=\sqrt{2}-2$