1

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?

José Carlos Santos
  • 427,504
James Warthington
  • 3,187

1 Answers1

5

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 $(-\infty,0]$

lab bhattacharjee
  • 274,582
  • Actually $f(y)$ is not increasing over the reals – zar Jun 06 '18 at 09:43
  • @zar, Sorry could not follow your statement – lab bhattacharjee Jun 06 '18 at 09:45
  • 2
    @zar $f(y)$ is stricly increasing on $[0,\infty[$, hence there exists a unique $y$ such that $f(y)=1$. – InsideOut Jun 06 '18 at 09:46
  • 1
    I see, but wolphram alpha gives another solution: $x\approx-2.73$ – zar Jun 06 '18 at 09:49
  • 1
    (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 06 '18 at 09:55
  • Not super relevant, but the intervals are wrong, aren't they? $0$ is not a root of $f'(y)=\mathrm{log}(3)3^y - \mathrm{log}(2)2^y$. – Piwi Jun 06 '18 at 10:28