Show that the equation x2−16x−16a=0(x>0, x+a>0)x^2 - 16x - 16a = 0\quad (x>0,\ x+a>0)x2−16x−16a=0(x>0, x+a>0), can be expressed in the form 2log2x=log2(x+a)+42\log_2 x = \log_2(x + a) + 42log2x=log2(x+a)+4
Given the equation x2−16x−16a=0x^2 - 16x - 16a = 0x2−16x−16a=0 has only one real root find the value of aaa.
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.