 
The goal here is to end up with loga = loga, so we can drop the logs. example 1: 2·log6 √(x) - log6(6x - 1) = 0 2·log6 √(x) = log6(6x - 1) Rearrange to eliminate the zero log6 (√(x))2 = log6(6x - 1) Reverse the power rule to bring the exponent inside the log We now have log = log. Drop the logs (√(x))2 = 6x - 1 x = 6x - 1 Simplify and solve - 5x = - 1 x = 1/5 or 0.2 Check: 2·log6 √(x) - log6(6x - 1) = 0 2·log6 √(0.2) - log6(6(0.2) - 1) = 0 Since none of the terms will be a log of zero or a log of a negative number, x = 0.2 is a solution example 2: logx + log(x - 1) = log(3x + 12) log[x(x - 1)] = log(3x + 12) Combine the first two terms by reversing the product rule x(x - 1) = 3x + 12 Drop the logs x2 - x = 3x + 12 Simplify and solve x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 x = 6 or x = - 2 Check both: Check x = 6 log6 + log(6 - 1) = log(3(6) + 12) Since none of the log arguments are zero or negative, x = 6 is a solution Check x = - 2 log(-2) + log(-2 - 1) = log(3(-2) + 12) Since two of the logs will be of negative numbers, x = -2 is NOT a solution The sole solution is x = 6 |