(71) \( \log x=1-\log (x-3) \)

To solve the equation \( \log x = 1 - \log (x-3) \), we can use the properties of logarithms. First, we can rewrite it as: \[ \log x + \log (x-3) = 1. \] Using the property that \( \log a + \log b = \log(ab) \), we have: \[ \log(x(x-3)) = 1. \] This implies: \[ x(x-3) = 10^1 = 10. \] Expanding the left-hand side gives us: \[ x^2 - 3x - 10 = 0. \] Now, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2}. \] This simplifies to: \[ x = \frac{3 \pm 7}{2}. \] Calculating the two possible values, we find: \[ x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-4}{2} = -2. \] However, since the logarithm is only defined for positive values, we discard \( x = -2 \). Therefore, the solution to the equation is: \[ \boxed{5}. \] It's always important to check that the solution we found satisfies the original logarithmic equations! For \( x = 5 \): \(\log(5) = 1 - \log(2)\) simplifies correctly, confirming that \( x = 5 \) is indeed valid.