How do I solve this log equation for x? $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$

Question

$3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$

I only managed to solve up to this step: $\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot \:125$

However, I'm not too sure about squaring both sides as I would have to handle ridiculously large numbers to find x, so I'm not too sure that I'm doing it correctly.

Answer 1

$$ (\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{4-x^2}\\ (\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{2-x}\sqrt{2+x}\\ (\sqrt{2+x})^2=125\\ 2+x=125\\ x=123. $$

Note that you have the following constraints on $x$: $$ \sqrt{2+x}>0\implies 2+x>0\implies x>-2\\ \sqrt{2-x}>0\implies 2-x>0\implies x<2 $$

If the original equation has a solution, it should be a value between the numbers $-2$ and $2$. The answer $123$ that we found does not belong to that interval. So, it looks like the original equation does not have solutions in real numbers.

Answer 2

You are almost there.

$$\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot125$$

$$\implies (2+x)\sqrt{2+x}\cdot\sqrt{2-x} = 125\sqrt{(2+x)(2-x)}$$

Here, notice that $x = 2$ and $x = -2$ are not the solutions for the original equation because $\log_50$ is not defined. Therefore, we can divide both sides by $\sqrt{2+x}\sqrt{2-x}$. Then, we have

$$2+x = 125 \implies x = 123$$

But this solution makes $\sqrt{2-x}$ a non-real number. Therefore this equation has no real solution.