Mathematics 2015 JAMB Past Questions - Question 42: If log10(x-3)+log10(x-2)=log10(2x2-8),Find the value of x

Correct Answer

D

Explanation

log10(x-3)+log10(x-2)=log10(2x2-8)=log10(x-3)(x-2)=log10(2x2-8)x2-5x+6=2x2-8x2+5x-14=0(x+7)(x-2)=0x=-7 or +2To solve the given equation, we can use the properties of logarithms. First, we can combine the logarithms on the left-hand side using the product rule of logarithms, which states that \(\log(a) + \log(b) = \log(a \cdot b)\).So, we have:\(\log_{10}(x-3) + \log_{10}(x-2) = \log_{10}(2x^2-8)\)Using the product rule, we can combine the logarithms on the left-hand side:\(\log_{10}((x-3)(x-2)) = \log_{10}(2x^2-8)\)Now, we can drop the logarithms and solve for x:\((x-3)(x-2) = 2x^2-8\)Expanding and simplifying the equation:\(x^2 - 5x + 6 = 2x^2 - 8\)\(0 = 2x^2 - x^2 - 5x + 6 + 8\)\(0 = x^2 - 5x + 14\)Now, we can solve for x by factoring or using the quadratic formula. Factoring the quadratic equation gives us:\((x-2)(x-7) = 0\)So, the solutions are \(x = 2\) and \(x = 7\). However, we need to check if these solutions satisfy the original equation. When \(x = 2\):\(\log_{10}(2-3) + \log_{10}(2-2) = \log_{10}(2 \cdot 2^2-8)\)\(\log_{10}(-1) + \log_{10}(0)\) is undefined, so \(x = 2\) is not a valid solution.When \(x = 7\):\(\log_{10}(7-3) + \log_{10}(7-2) = \log_{10}(2 \cdot 7^2-8)\)\(\log_{10}(4) + \log_{10}(5) = \log_{10}(98)\)\(\log_{10}(20) = \log_{10}(98)\)Since \(\log_{10}(20) \neq \log_{10}(98)\), the solution \(x = 7\) is also not valid.Therefore, there is no valid solution for x that satisfies the original equation.