Logarithm Problem

First start with changing the base of log x: $log(x) = \frac{ln(x)} {ln(10)}$ . The equation can be rewritten as: $(ln(x))^2 - \frac{ln(x)} {ln(10)} = ln(x) (ln(x) - \frac{1} {ln(10)}) =0$ . So ln (x) =0, which gives x = 1. Or $ln(x) - \frac{1} {ln(10)} = 0$ . This leads to $x = e^{ \frac{1} {ln(10)}}$ .