Additional Mathematics 0606 / Logarithms and Exponents / 2.112.11: Equation in lg x (repeat)0606/22/O/N/16 — Question 3 · 5 marks · Log laws, EquationsMark as done · Save for later · Show all solutionsIdentical to Question 2.3 — the same question appeared on paper 22.Solve the equation 2lgx−lg(x+102)=12\lg x - \lg\left(\dfrac{x + 10}{2}\right) = 12lgx−lg(2x+10)=1.[5]▸ Answer← 2.10: Index equation; solve log a b minus half equals log b a2.12: Equation with fractional indices →
