Using y=2xy = 2^xy=2x as a substitution, show that 4x−2x+2−5=04^x - 2^{x+2} - 5 = 04x−2x+2−5=0 can be written as y2−4y−5=0y^2 - 4y - 5 = 0y2−4y−5=0.
Hence, show that the equation 4x−2x+2−5=04^x - 2^{x+2} - 5 = 04x−2x+2−5=0 has x=log25x = \log_2 5x=log25 as its only solution.
Practise AQA A Level Maths 1.9 F: Exponentials and logarithms with exam-style questions for A Level Maths. 37 questions covering 1.9.1 Exponential functions and their graphs, 1.9.2 Gradient of e^kx (A-level only), 1.9.3 Logarithms as inverse functions, 1.9.4 Laws of logarithms, 1.9.5 Solving exponential equations, 1.9.6 Logarithmic graphs to estimate parameters (A-level only), and 1.9.7 Exponential growth and decay, matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
