solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$

Question

Solve for $x$

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

Answer 1

Assume that $\log$ is the natural logarithm and $x > 1$ (Otherwise, it would not make sense the power in LHS). Since $\log \sqrt{x} = \dfrac{\log x}{2}$, we can simplify the equation as $$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} \implies \left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 3)} = 1$$ $$\implies \left(\log^2x + 2\log x - 3\right)\log\left(\dfrac{\log x}{2}\right) = 0.$$

Thus, $$\log\left(\dfrac{\log x}{2}\right) = 0 \implies x = e^{2}.$$ Or, $$\log^2x + 2\log x - 3 = 0 \implies \log x = -1\pm2 \implies x = e ~~\text{and}~~x = e^{-3}.$$

Answer 2

$\frac{\log x}{2}=\log{\sqrt{x}}$

Using this,

$( \log{\sqrt{x}})^{(\log x)^2+2\log x-2}= \log{\sqrt{x}}$

$\implies (\log x)^2+2\log x-2=1$

$\implies (\log x+3)(\log x-1)=0$

$\implies x=e^{-3}$ or $e$

Again,

$\log{\sqrt{x}}=1$ gives a solution.

$\implies x=e^2$ is also a solution

Answer 3

Let's set $\log x=u$ and therefore $\log \sqrt x=\frac{u}{2}$. The equation becomes $\left (\frac{u}{2}\right )^{u^2+2u-2}=\frac{u}{2}$ dividing both sides by $\frac{u}{2}$ the equation is $\left (\frac{u}{2}\right )^{u^2+2u-3}=1$ this will occur when 1) the exponent is 0 (provided that it's base is not also 0), 2) when $\frac{u}{2}=\pm 1$ and the negative case is only true if and only if the exponent is even. So 1) is a simple quadratic with roots -3 and 1. Then 2) happens when $u=\pm 2$ checking the negative and the exponent is odd (so it is not a solution) so we have that $u=\log x=-3, 1, 2\implies x=e^{-3},e,e^{2}$ that case of the -1 is important to remember for these types of questions that's why I brought it up even though it is not applicable for this one.