Theory question 110
Indices and Logarithms(a) Solve 23y+2 − 7(22y+2) − 31(2y) − 8 = 0, y ∈ R.
(b) Find ∫(√(x2 + 1))x dx
Show the model answer
Model answer
(a) Let p = 2ʸ: 4p³ − 28p² − 31p − 8 = 0.
p = 8 is a root: (p − 8)(4p² + 4p + 1) = (p − 8)(2p + 1)² = 0.
2ʸ > 0, so p = 8 and y = 3.
(b) Let u = x² + 1, du = 2x dx: ∫ ½√u du = ⅓u^(3/2) + c = ⅓(x² + 1)^(3/2) + c.
Explanation
A substitution turns (a) into a cubic in p = 2ʸ.