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Solve the eqation for all real $x$: $\log_2(x^2+7)+\log_3(x+6)=6$.

What I tried: $\log_2(x^2+7)=a$ and $\log_3(x+6)=b$, then $a+b=6$ and $2^a=3^{2b}-4\cdot3^{b+1}+43$. But the problem is $a$ and $b$ don't have to be integers...

It is easy to show that for $x>3$ and $3>x\ge 0$ there are no soultions. Which leaves us only with the interval $0>x>-6$.

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Take derivative and consider where left part increases and decreases, where are extrema (at least approximately). After proving that it has a single root, we can guess that it is $x=3$.

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Hint: $x$ is a natural number. Also $log_2(x^2+7)$ and $log_3(x+6)$ are natural numbers. You will end up with $4+2=6$

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  • $\begingroup$ Why are $x$ and the logarithms natural numbers? $\endgroup$ – user26486 Feb 9 '15 at 19:38
  • $\begingroup$ Well, if thats it what would be easy. But why does x has to be a natural number and the other two too? $\endgroup$ – HeatTheIce Feb 9 '15 at 19:39
  • $\begingroup$ $x$ is really a natural number. $3^2+7=16=2^4,3+6=9=3^2$. $\endgroup$ – kryomaxim Feb 9 '15 at 19:42
  • $\begingroup$ @kryomaxim There exists a solution that is a natural number, but you haven't proved that there couldn't be non-natural solutions. $\endgroup$ – user26486 Feb 9 '15 at 20:07
  • $\begingroup$ mathe-fa.de/de#result; "English"; plot the function for the right Hand side (a constant function) and another for the left Hand side and you will see that there is only 1 solution (I have checked that on mathe-fa.de). $\endgroup$ – kryomaxim Feb 9 '15 at 20:10

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