Quantitative Aptitude

Taking $\log_{10}$ on both sides: $(\log_{10} x)^2 = \log_{10}(100x) = 2 + \log_{10} x$. Let $y = \log_{10} x$. $y^2 - y - 2 = 0 \implies (y - 2)(y + 1) = 0$. For $x > 1$, $y = 2 \implies x = 10^2 = 100$.