# 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