**Problem 9:** Solve the equation: \[ \log x + \log(x - 21) = 2 \] **Solution:** To solve this equation, use the properties of logarithms. The sum of logs can be converted to the log of a product: \[ \log(x) + \log(x - 21) = \log(x(x - 21)) \] This simplifies to: \[ \log(x^2 - 21x) = 2 \] Rewrite the equation in exponential form: \[ x^2 - 21x = 10^2 \] \[ x^2 - 21x = 100 \] Now, rearrange the equation to set it to zero: \[ x^2 - 21x - 100 = 0 \] Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \(a = 1\), \(b = -21\), and \(c = -100\): \[ x = \frac{-(-21) \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot (-100)}}{2 \cdot 1} \] \[ x = \frac{21 \pm \sqrt{441 + 400}}{2} \] \[ x = \frac{21 \pm \sqrt{841}}{2} \] \[ x = \frac{21 \pm 29}{2} \] Calculate the two potential solutions: \[ x = \frac{21 + 29}{2} = \frac{50}{2} = 25 \] \[ x = \frac{21 - 29}{2} = \frac{-8}{2} = -4 \] Since logarithms are only defined for positive values, discard \(x = -4\). Thus, the solution is: \[ x = 25 \]