14. Solve for \( x \) (a) \( \log x+\log (x-3)=1 \) (b) \( 3^{1-x}=2 \) 15. Prove that \( \log _{\sqrt{b}}=2 \log _{b} x \)

To solve \( \log x + \log (x - 3) = 1 \), you can combine the logs into one using the product rule of logarithms: \( \log [x(x - 3)] = 1 \). This simplifies to \( x^2 - 3x = 10 \) after exponentiating, leading to \( x^2 - 3x - 10 = 0 \). Applying the quadratic formula gives you \( x = 5 \) and \( x = -2 \). Since \( x \) must be greater than 3, the only solution is \( x = 5 \). For the second equation, \( 3^{1-x} = 2 \), take the logarithm of both sides. This gives you \( 1 - x = \log_3 2 \), leading to \( x = 1 - \log_3 2 \). Convert to a natural log format if desired—remember that \( \log_3 2 = \frac{\log 2}{\log 3} \) for those fun logarithmic transformations! Now, to prove \( \log_{\sqrt{b}} x = 2 \log_{b} x \), recall that \( \log_{\sqrt{b}} x \) can be expressed as \( \frac{\log_b x}{\log_b \sqrt{b}} \). Since \( \sqrt{b} = b^{1/2} \), we have \( \log_b \sqrt{b} = \frac{1}{2} \). Plugging this back in gives you \( \frac{\log_b x}{1/2} = 2 \log_b x \), completing the proof in a flash!