Additional Mathematics 0606 / Logarithms and Exponents / 1.31.3: Logarithmic equations with bases 3, 9 and 40606/11/M/J/19 — Question 7 · 8 marks · Log laws, Change of base, EquationsMark as done · Save for later · Show all solutions(a)Solve log3x+log9x=12\log_{3} x + \log_{9} x = 12log3x+log9x=12.[3]▸ Answer(b)Solve log4(3y2−10)=2log4(y−1)+12\log_{4}(3y^{2} - 10) = 2\log_{4}(y - 1) + \dfrac{1}{2}log4(3y2−10)=2log4(y−1)+21.[5]▸ Answer← 1.2: Single logarithm and a quadratic in log a 51.4: Quadratic in log base 3 by substitution →
