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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?

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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 $(-\infty,0]$

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  • $\begingroup$ Actually $f(y)$ is not increasing over the reals $\endgroup$ – zar Jun 6 '18 at 9:43
  • $\begingroup$ @zar, Sorry could not follow your statement $\endgroup$ – lab bhattacharjee Jun 6 '18 at 9:45
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    $\begingroup$ @zar $f(y)$ is stricly increasing on $[0,\infty[$, hence there exists a unique $y$ such that $f(y)=1$. $\endgroup$ – InsideOut Jun 6 '18 at 9:46
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    $\begingroup$ I see, but wolphram alpha gives another solution: $x\approx-2.73$ $\endgroup$ – zar Jun 6 '18 at 9:49
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    $\begingroup$ (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) $\endgroup$ – zar Jun 6 '18 at 9:55

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