Exponent equation with common power Solve for $x$ in
$$\log_{2}(2^{x-1}+3^{x+1}) = 2x-\log_{2}(3^x)$$
I simplified by doing:
$$\log_{2}(3^x \cdot 2^{x-1} + 3^{2x+1}) = 2x$$
$$\frac{6^x}{2} + 3^{2x+1} = 2^{2x}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$
Where do I go from here? I tried moving more numbers around, but I haven't been able to get any closer to solving for x. Any help is appreciated.

Put $3^x=a$ and $2^x=b$. Then
$$6a^2+ab-2b^2=0$$
$$\Rightarrow 6a^2+4ab-3ab-2b^2=0$$
$$\Rightarrow (3a+2b)(2a-b)=0$$
Now $3a+2b$ is always positive. So $2a-b=0$.
Therefore, $$2\cdot 3^x = 2^x$$
$$\Rightarrow x=\log_{2/3} 2=\frac{1}{\log_2 (2/3)}$$
