Logarithm Evaluation Equation

1. **Problem 1:** Evaluate \( \frac{3}{8.496 \times 2.41} \times 3941 \) using logarithm tables. 2. **Problem 2:** Solve the equation \( \frac{3}{x-2} = \frac{4}{x-3} + 2 \). --- ### Problem 1: Using logarithm tables to evaluate \( \frac{3}{8.496 \times 2.41} \times 3941 \) 1. Write the expression clearly: $$ \frac{3}{8.496 \times 2.41} \times 3941 $$ 2. Calculate the product in the denominator: $$ 8.496 \times 2.41 $$ 3. Using logarithm tables, find \( \log(8.496) \) and \( \log(2.41) \), then add them: $$ \log(8.496) + \log(2.41) = \log(8.496 \times 2.41) $$ 4. Find \( \log(3) \) and \( \log(3941) \). 5. The expression can be rewritten using logarithms as: $$ \log\left( \frac{3}{8.496 \times 2.41} \times 3941 \right) = \log(3) + \log(3941) - \log(8.496) - \log(2.41) $$ 6. Calculate the sum and difference of these logarithms. 7. Use the antilogarithm (inverse log) to find the final value. --- ### Problem 2: Solve \( \frac{3}{x-2} = \frac{4}{x-3} + 2 \) 1. Write the equation: $$ \frac{3}{x-2} = \frac{4}{x-3} + 2 $$ 2. Bring all terms to one side: $$ \frac{3}{x-2} - \frac{4}{x-3} = 2 $$ 3. Find a common denominator \( (x-2)(x-3) \) and combine the fractions: $$ \frac{3(x-3) - 4(x-2)}{(x-2)(x-3)} = 2 $$ 4. Simplify the numerator: $$ 3x - 9 - 4x + 8 = -x - 1 $$ 5. The equation becomes: $$ \frac{-x - 1}{(x-2)(x-3)} = 2 $$ 6. Multiply both sides by \( (x-2)(x-3) \): $$ -x - 1 = 2(x-2)(x-3) $$ 7. Expand the right side: $$ 2(x^2 - 5x + 6) = 2x^2 - 10x + 12 $$ 8. Write the equation: $$ -x - 1 = 2x^2 - 10x + 12 $$ 9. Bring all terms to one side: $$ 0 = 2x^2 - 10x + 12 + x + 1 $$ $$ 0 = 2x^2 - 9x + 13 $$ 10. Solve the quadratic equation: $$ 2x^2 - 9x + 13 = 0 $$ 11. Use the quadratic formula: $$ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 2 \times 13}}{2 \times 2} = \frac{9 \pm \sqrt{81 - 104}}{4} = \frac{9 \pm \sqrt{-23}}{4} $$ 12. Since the discriminant is negative, there are no real solutions. --- **Final answers:** - Problem 1: Use logarithm tables to find the numerical value as explained. - Problem 2: No real solution because the discriminant is negative.