Question

Solve : $\log_{10}^2 x + \log_{10}x^2 = \log_{10} 3-1$

Answer 1

The solutions for $x$ are: $x = 10^{-1 + \sqrt{\log_{10} 3}}$ and $x = 10^{-1 - \sqrt{\log_{10} 3}}$

Answer 2

The given equation is $\log_{10}^2 x + \log_{10}x^2 = \log_{10} 3-1$. Using logarithm properties, this becomes $(\log_{10} x)^2 + 2\log_{10} x = \log_{10}(3/10)$. Let $y = \log_{10} x$. The equation is $y^2 + 2y - \log_{10}(3/10) = 0$. Solving for $y$ using the quadratic formula gives $y = -1 \pm \sqrt{1 - \log_{10}(3/10)} = -1 \pm \sqrt{\log_{10}3}$. Thus, $\log_{10} x = -1 + \sqrt{\log_{10}3}$ or $\log_{10} x = -1 - \sqrt{\log_{10}3}$. Converting to exponential form, $x = 10^{-1 + \sqrt{\log_{10}3}}$ or $x = 10^{-1 - \sqrt{\log_{10}3}}$.