How do you solve the equation #log_9 (x-3)+log_9(3x-2)=log_9 18#?
1 Answer
Jul 6, 2017
# x = 4.54647 # to 5 dp
Explanation:
We will need a to use several of the properties of logarithms, namely
# log A + log B -= log AB #
# log A = log B iff A=B #
We have:
# log_9 (x-3)+log_9(3x-2)=log_9 18 #
# :. log_9 (x-3)(3x-2)=log_9 18 #
# :. (x-3)(3x-2)=18 #
# :. 3x^2-11x+6=18 #
# :. 3x^2-11x-12=0 #
We can solve this quadratic by completing the square or using the quadratic formula, and we get:
# x=11/6+-sqrt(265)/6 #
# \ \ = -0.87980, 4.54647 # to 5 dp
Note that we should reject any solution that could have yielded a negative argument to the initial logarithm question
# x-3 \ gt 0 \ \=> x gt 3 #
# 3x-2 gt 0 => x gt 2/3 #
Yielding the only valid solution;
# x = 4.54647 # to 5 dp
