Logarithmic Equation

Simplify the log terms: $5 - \frac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1-x^2)$ Notice that $\log_{10}(1-x^2) = \log_{10}(1+x) + \log_{10}(1-x)$. So, $5 - \frac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1+x) - \frac{1}{2}\log_{10}(1-x)$

Cancel $-\frac{1}{2}\log_{10}(1+x)$ from both sides: $5 + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1-x)$ $5 = -\frac{5}{2}\log_{10}(1-x)$ $\log_{10}(1-x) = -2$ $1 - x = 10^{-2} = 0.01$ $x = 0.99$ Therefore, $100x = 99$.