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

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 $(-\infty,0]$
