Additional Mathematics 0606 / Logarithms and Exponents / 2.32.3: Equation in lg x0606/21/O/N/16 — Question 3 · 5 marks · Log laws, EquationsMark as done · Save for later · Show all solutionsSolve the equation 2lgx−lg(x+102)=12\lg x - \lg\left(\dfrac{x + 10}{2}\right) = 12lgx−lg(2x+10)=1.[5]▸ Answer← 2.2: Binomial expansion applied to an exponential equation2.4: Log equation with base 5 →