EXPONENTIAL AND LOGARITHMIC FUNCTIONS

e a x + b = p ln ( a x + b ) = q ln A = p e p = A We shall look at how to solve equations of the forms: e a x + b = p ln ( a x + b ) = q and, Firstly it is essential to know the relationship between a logarithmic and an exponential statement: i.e. If: ln A = p Then: e p = A … and vice versa.

Example 1: Solve the equation 7 + e 2x + 1 = 50 ln A = p e p = A ln 43 – 1 2 ln 43 = 2x + 1 x = = 1.38 ( 3 sig.figs.) Example 2: Solve the equation 1 + 2 ln ( x – 5 ) = 7 1 + 2 ln ( x – 5 ) = 7 2 ln ( x – 5 ) = 6 ln ( x – 5 ) = 3 ln A = p e p = A e3 = x – 5 x = 5 + e3 = 25.1 ( 3 sig.figs.)

Example 3: The temperature, θ° C, of a cup of coffee at time t minutes after being made is given by θ = 18 + 64 e – 0.04 t. a) Find the initial temperature of the coffee. b) Find how long it takes for the temperature of the coffee to be 50 ° C. a) θ = 18 + 64 e – 0.04 t Initially, t = 0 θ = 18 + 64 e 0 = 18 + 64 = 82 The temperature initially is 82° C. b) When θ = 50, 50 = 18 + 64 e – 0.04 t 32 = 64 e – 0.04 t 0.5 = e – 0.04 t ln A = p e p = A ln 0.5 = – 0.04 t t = 17.3 i.e. approximately 17 minutes.

ln A = p e p = A e q = ax + b ax + b = ln p If: Then: Summary of key points: If: ln A = p Then: e p = A Hence to solve an equation of the form: ln (a x + b) = q We have: e q = ax + b and to solve an equation of the form: e ax + b = p We have: ax + b = ln p This PowerPoint produced by R.Collins ; Updated Mar.2010