Solve $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$ for $x\in\mathbb{R}.$ Solve $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$ for $x\in\mathbb{R}.$ There is a brute force method that relies on breaking each exponential up into $2^x$’s and $3^x$’s and substituting, but I am looking for a more elegant one line or so proof (not brute force). Any suggestions? 

Here's the way I did it: $$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x)^3+(3^x)^3}{(3\cdot2^2)^x+(2\cdot3^2)^x}=\frac{(2^x)^3+(3^x)^3}{3^x\cdot(2^x)^2+2^x\cdot(3^x)^2}\tag1$$
Let $a=2^x, b=3^x$ so that $(1)$ is equivalent to $$\frac{a^3+b^3}{ba^2+ab^2}=\frac{(a+b)(a^2-ab+b^2)}{(a+b)(ab)}=\frac{7}{6}\tag2$$
Then cancelling the $a+b$ term and cross multiplying (we are allowed to do so since $a\neq -
b \text{ for all } x)$ yields $$6a^2-6ab+6b^2=7ab\iff6a^2-13ab+6b^2=0\tag3$$ Then $(3)$ yields the nice simplifications $$a=\frac{2b}{3}\text{ or } a=\frac{3b}{2}\implies 2^x=\frac{2\cdot3^x}{3}\text { or  } 2^x=\frac{3\cdot 3^x}{2}\tag4$$
Thus $(4)$ gives $$2^{x-1}=3^{x-1}\text{ or }\space\space 2^{x+1}=3^{x+1}\iff \boxed{x=1\text{ or } x=-1}$$
