How do you solve for x in #log_ 6 (2-x) + log _ 6 (3-4x) = 1#?

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Answer 1 · 2016-02-06T05:47:47

#x= 0#

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#

#x= 0 # or #x= 5#

Always check your answer if you have 2 solutions when solving a log equation .

#color(blue)(Check " " x= 5)#

#x= 5#

#log_6(2-5) + log_6(3-4(5))= 1#

#log_6(-3) + log_6(-17) = 1#

#color(red)( x = 5 " Extraneous " solution)#

Can't evaluate the expression on the left, because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction

#color(blue)(Check " " x= 0)#

#color(red)(log_6(2-0)+log_6(3-4(0))= 1#

#log_6(2)+log_6(3) = 1#

#log_6(6) = 1#

# 1= 1#

Solution #color(red)(x= 0)#