[Algebra]
Problem: Determine all real numbers $x$ for which $2 \log _{2}(x-1)=1-\log _{2}(x+2)$.

Solution: Using logarithm and exponent laws, we obtain the following equivalent equations:

$$
\begin{aligned}
2 \log _{2}(x-1) & =1-\log _{2}(x+2) \\
2 \log _{2}(x-1)+\log _{2}(x+2) & =1 \\
\log _{2}\left((x-1)^{2}\right)+\log _{2}(x+2) & =1 \\
\log _{2}\left((x-1)^{2}(x+2)\right) & =1 \\
(x-1)^{2}(x+2) & =2^{1} \\
\left(x^{2}-2 x+1\right)(x+2) & =2 \\
x^{3}-3 x+2 & =2 \\
x^{3}-3 x & =0 \\
x\left(x^{2}-3\right) & =0
\end{aligned}
$$

and so $x=0$ or $x=\sqrt{3}$ or $x=-\sqrt{3}$.

Note that if $x=0$, then $x-1=-1<0$ and so $\log _{2}(x-1)$ is not defined. Thus, $x \neq 0$. Note that if $x=-\sqrt{3}$, then $x-1=-\sqrt{3}-1<0$ and so $\log _{2}(x-1)$ is not defined. Thus, $x \neq-\sqrt{3}$.

If $x=\sqrt{3}$, we can verify that both logarithms in the original equation are defined and that the original equation is true. We could convince ourselves of this with a calculator or we could algebraically verify that raising 2 to the power of both sides gives the same number, so the expressions must actually be equal.

Therefore, $x=\sqrt{3}$ is the only solution.

Final answer: $\sqrt{3}$