Solve $\log_3(x^2+2x+1)=\log_2(x^2+2x)$

Question

Solve $\log_3(x^2+2x+1)=\log_2(x^2+2x)$

I have tried to do to as followed:

$\log_3(x^2+2x+1)=\frac{\log_3(x^2+2x)}{\log_3(2)}$

$\iff\log_3(x^2+2x+1).\log_3(2)=\log_3(x^2+2x)$

Is it possible to proceed this way? Or should one approach this differently?

Answer 1

If $\log_3(x^2+2x+1)=\log_2(x^2+2x)=y$

$f(y)=3^y-2^y=1$

Now $f(y)$ is an increasing function in $[0,\infty)$ and decreasing in $(-\infty,0]$