Using y=2xy = 2^xy=2x as a substitution, show that 4x−2x+1−2=04^x - 2^{x+1} - 2 = 04x−2x+1−2=0 can be written as y2−2y−2=0y^2 - 2y - 2 = 0y2−2y−2=0.
Hence, show that the equation 4x−2x+1−2=04^x - 2^{x+1} - 2 = 04x−2x+1−2=0 has x=log2(1+3)x = \log_2(1+\sqrt{3})x=log2(1+3) 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.