exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$

Let $a=3^x$ and $b=2^x$ and $c=5^x$.

Then we have that $$ab+b^3+ac=a^2+bc+ab^2$$ $$ab+b^3+ac-a^2-bc-ab^2=0$$ $$a(b-a)+b^2(b-a)-c(b-a)=0$$ $$(b-a)(a+b^2-c)=0$$

Now $a=b \implies x=0$ and $a+b^2-c=0 \implies 3^x+4^x=5^x$ which gives solution for $x=2$ only and not for any higher integer $x$ by Fermat's Last Theorem.

So these are the $2$ solutions.

Hint: your equation can be factorized as $$\left(2^x-3^x\right) \left(2^{2 x}+3^x-5^x\right)=0$$