Find all values of $x$

Question

Determine all real values of $x$ such that: $$\log_{2}(2^{x-1} + 3^{x+1}) = 2x - \log_{2}(3^x) $$

Let $u = 2^x$ and let $y = 3^x$

For ease, let $\log_{2}$ be represented by just $\log$ so:

Then, $\log(u/2 + 3y) = \log(u^2) - \log(y)$, which means, $\log(u/2 + 3y) = \log(u^2/y)$ and so:

$u/2 + 3y = u^2/y \implies \frac{uy}{2} + 3y^2 - u^2 = 0$

Factoring a little, $y(u/4 + 3y) + u(y/4 - u) = 0$ Doesn't help, replace back:

$2^{x-1}3^x + 3^{2x + 1} - 2^{2x} = 0$. But this is all I can go upto? A hint?

Answer 1

Hint: Your last equation is simply $(u-2y)(u+1.5y)=0$. Since $u,y>0$, we get $u=2y$. See my comments for more information.

Answer 2

the main idea is to write $2x$ as $$\log_2 2^{2x}$$