$\frac{8^x-2^x}{6^x-3^x} = 2$
I was able to use difference of cubes on the numerator, and setting $a= 2^x$ would give the following:
$\frac{a(a-1)(a+1)}{3^x(a-1)} = 2$
Now we have a common factor and with $a \not= 1$. Then I moved on to express the $3^x$ term in terms of $2^x$ to get like terms on both sides. This would yield:
$a(a+1) = 2(3^x)$
$a(a+1) = 2(2^{(log_2(3))})^x$
$a(a+1) = 2(a^{(log_2(3))})$
$a+1 = 2a^{(log_2(\frac{3}{2}))}$
Here is where I am stuck, is there an algebraic way to continue this method until we get the answer?
Thank you for your help.