Don't overlook the base!!

$\frac{\log x}{2}$ is the same as $\log \sqrt{x}$ , which implies that $\log^2 x + \log x^2 - 2 = 1$ . On solving this, you will get 2 values of $x$ .

But, when $\frac{\log x}{2} =1$ , then no matter what the value of $\log^2 x + \log x^2 - 2$ is, the above equation will always be true. This is because 1 raised to anything is 1. For this condition, we get another value of $x$ . So, in total 3 values of $x$ satisfy the equation.

second time my answer is right