$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$

Question

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?

Answer 1

$$0=2^{x+3}(2^{2x-10}-1)+3^{x-5}(3^{x-5}-1)=f(x)$$

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 $x = 5$.

Answer 2

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, $$\begin{align*} 2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10} &\Rightarrow \begin{cases} 2^{x+3} = 2^{3x-7} \\ 3^{x-5} = 3^{2x-10} \end{cases} \\ &\Rightarrow \begin{cases} x+3 = 3x-7 \\ x-5 = 2x-10 \end{cases} \\ &\Rightarrow \begin{cases} 2x = 10 \\ x = 5 \end{cases}\\ &\Rightarrow x=5 \end{align*}$$