 
The goal is to end up with log = constant, and 'complete the circle' to get an exponential equation. example 1: log4(2x + 4) = 2 42 = 2x + 4 Complete the circle 16 = 2x + 4 Simplify and solve 12 = 2x 6 = x Check: log4(2(6) + 4) = 2 The argument of the log is neither zero nor negative, so x = 6 is a solution This is also simple enough to evaluate to show it works: log4(2(6) + 4) = 2 log416 = 2 2 = 2 example 2: log4x = 1 - log4(x - 3) log4x + log4(x - 3) = 1 Rearrange so both logs are on one side, and the constant on the other log4(x(x - 3)) = 1 Combine the logs by reversing the product rule 41 = x(x - 3) Complete the circle, simplify and solve 4 = x2 - 3x 0 = x2 - 3x - 4 (x - 4)(x + 1) = 0 x = 4 or x = - 1 Check both: check x = 4 log4x = 1 - log4(x - 3) log44 = 1 - log4(4 - 3) Neither log argument is zero or negative, so x = 4 is a solution check x = - 1 log4x = 1 - log4(x - 3) log4(-1) = 1 - log4(-1 - 3) Both arguments are negative, so x = -1 is NOT a solution The sole solution is x = 4 example 3: log2(x2 - 6x) = 3 + log2(1 - x) log2(x2 - 6x) - log2(1 - x) = 3 Rearrange with logs on one side and the constant on the other log2[(x2 - 6x) / (1 - x)] = 3 Combine logs by reversing the quotient rule 23 = (x2 - 6x) Complete the circle, simplify and solve (1 - x) 8 = (x2 - 6x) (1 - x) 8(1 - x) = x2 - 6x 8 - 8x = x2 - 6x x2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = - 4 or x = 2 Check both: check x = - 4 log2((-4)2 - 6(-4)) = 3 + log2(1 - -4) Both arguments are positive, so x = - 4 is a solution check x = 2 log2(x2 - 6x) = 3 + log2(1 - x) log2(22 - 6(2)) = 3 + log2(1 - 2) Both arguments are negative, so x = 2 is NOT a solution The sole solution is x = - 4 NOTE: The quadratic equation you must solve could have fractional solutions, requiring factoring by decomposition. It is also possible that the roots could be irrational or non-existant, which would require the use of the General Quadratic Formula. |