# 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 $(-\infty,0]$
• Actually $f(y)$ is not increasing over the reals – zar Jun 6 '18 at 9:43
• @zar $f(y)$ is stricly increasing on $[0,\infty[$, hence there exists a unique $y$ such that $f(y)=1$. – InsideOut Jun 6 '18 at 9:46
• I see, but wolphram alpha gives another solution: $x\approx-2.73$ – zar Jun 6 '18 at 9:49
• (Ok, my mistake, you are talking about the unique $y$ such that $f(y)=1$, and for that particular $y$ there exist two different $x$ which are solutions of the original equation) – zar Jun 6 '18 at 9:55