Additional Mathematics 0606 / Indices and surds / 1.111.11: Exponential equation and index laws0606/13/M/J/16 — Question 2 · 5 marksMark as done · Save for later · Show all solutions(a)Solve the equation 163x−1=8x+216^{3x-1} = 8^{x+2}163x−1=8x+2.[3]▸ Answer(b)Given that (a13b−12)3a−23b12=apbq\dfrac{(a^{\frac{1}{3}}b^{-\frac{1}{2}})^{3}}{a^{-\frac{2}{3}}b^{\frac{1}{2}}} = a^{p}b^{q}a−32b21(a31b−21)3=apbq, find the value of each of the constants ppp and qqq.[2]▸ Answer▸ Official mark scheme← 1.1: Exponential equation and index laws2.1: Rationalising and expanding surd squares →
