How do you solve for x in #log_ 6 (2-x) + log _ 6 (3-4x) = 1#?
1 Answer
Explanation:
Property of Logarithmic expression
#log A + log B = Log(AB) " " " " " (1)#
#n log A= log A^n " " " " (2)#
#log_ay= x => a^x = y " " " (3) #
Given :
#log_6(2-x)+ log_6(3-4x) = 1#
Rewrite as:
Using rule (2)
#log_6(2-x)(3-x)= 1#
#log_6(6-2x-3x+x^2)= 1#
#log_6(6-5x+x^2) = 1#
Using rule (3)
#6^1 = 6-5x^2 + x^2#
Now we have a simple quadratic equation, let's solve it.
#x^2 -5x+6= 6#
#-6 " " " -6#
#x^2 -5x = 0#
#x(x-5) = 0#
Always check your answer if you have 2 solutions when solving a log equation .
#log_6(-3) + log_6(-17) = 1#
Can't evaluate the expression on the left, because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction
#log_6(2)+log_6(3) = 1#
#log_6(6) = 1#
# 1= 1#
Solution