<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] As @JohnHughes hinted at in the comments, the trick to solving $2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10}$ is in realizing that powers of two are always even, and powers of three are always odd, so a power of two won't ever affect the value of a power of three. (Edit: This is only true when you are multiplying powers of two by powers of three, not adding them, so this is one solution, there might be others.)
So, basically,
$$\begin{align*}
2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10}
&\Rightarrow \begin{cases}
2^{x+3} = 2^{3x-7} \\
3^{x-5} = 3^{2x-10}
\end{cases} \\
&\Rightarrow 
\begin{cases}
x+3 = 3x-7 \\
x-5 = 2x-10
\end{cases} \\
&\Rightarrow 
\begin{cases}
2x = 10 \\
x = 5
\end{cases}\\
&\Rightarrow x=5
\end{align*}$$
</s>