Solve the logarithmic equation by $x$ 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 $x&gt;3$ and $3&gt;x\ge 0$ there are no soultions. Which leaves us only with the interval $0&gt;x&gt;-6$.
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 $x=3$.
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 $4+2=6$