<s>[INST] $2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$ Related to solving $~~2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
I've tried some arithmetic to find something like 
$a^x=b~\Longrightarrow~x=\log_{a}b$
But what I've found is that
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
$2^8\cdot2^{x-5}+3^{x-5}=2^8\cdot2^{3(x-5)}+3^{2(x-5)}$
$256\cdot2^{x-5}+3^{x-5}=256\cdot(2^{x-5})^3+(3^{x-5})^2$
$256\cdot a +b=256\cdot a^3+b^2$
$256 \cdot a(a^2-1)+b(b-1)=0$
I don't know how to solve this equation above. Maybe it doesn't the better way to solve my question (tittle) or maybe it is wrong.
Can someone help me to solve that?
 [/INST] $$0=2^{x+3}(2^{2x-10}-1)+3^{x-5}(3^{x-5}-1)=f(x)$$
As $2^{x+3},3^{x-5}>0$ for real $x,$
If $x-5>0$, then $f(x)>0$, and if $x-5<0$, then $f(x)<0$.
Since we wanted $f(x) = 0$, it must be true that $x - 5 \not< 0$ and $x - 5 \not> 0$. So $x - 5 = 0$, and thus $x = 5$.
</s>