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.
77 exam-style questions on CCEA A Level Maths 1.5 Exponentials and logarithms, covering 1.5.1 Exponentials and logarithms, 1.5.2 Exponentials and logarithms, 1.5.3 Exponentials and logarithms, 1.5.4 Exponentials and logarithms, 1.5.5 Exponentials and logarithms, 1.5.6 Exponentials and logarithms, 1.5.7 Exponentials and logarithms, and 1.5.8 Exponentials and logarithms. Each one has a worked solution and a mark scheme showing where the marks go.