NMTC Inter Level Problem 5

$\log_{10} 100x = \log_{10} 100+\log_{10} x$

Let $\log_10 x =a$

Thus $a^2=a+2$

This is a quadratic with roots $a=2$ , $a=-1$

Thus $\log_{10} x=-1 \text{ or } 2 \implies x = \dfrac{1}{10} \text{ or } 100$ , they're $\boxed{2}$ solutions.

$(\log_{10} x)^2 =\log_{10}(100 x) =\log_{10} (100) +\log_{10} x= 2+\log_{10} x~~\implies\\ (\log_{10} x)^2- \log_{10} x-2=0$
This is a quadratic equation in variable $\log x$ . Hence two solutions.