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.
277 exam-style questions on Edexcel A Level Maths Exponentials and Logarithms, covering 14.1 Exponential Functions, 14.2 y = e^x, 14.3 Exponential Modelling, 14.4 Logarithms, 14.5 Laws of Logarithms, 14.6 Solving Equations using Logarithms, 14.7 Working with Natural Logarithms, and 14.8 Logarithms and Non-Linear Data. Each one has a worked solution and a mark scheme showing where the marks go.