How do you solve #log_3(x-6)=log_9(2x)#?

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Answer 1 · 2016-09-30T16:13:34

#x = 7+ sqrt(13)#

#log_3(x- 6) - log_9(2x) = 0#

#log(x - 6)/log3 - log(2x)/log9 = 0#

#log(x- 6)/log3 - log(2x)/(2log3) = 0#

Put on a common denominator:

#(2log(x - 6))/(2log3) - log(2x)/(2log3) = 0#

#log_9(x - 6)^2- log_9(2x) = 0#

#log_9((x^2 - 12x + 36)/(2x)) = 0#

#(x^2 - 12x + 36)/(2x) = 9^0#

#(x^2 - 12x + 36)/(2x) = 1#

#x^2 - 12x + 36 = 2x#

#x^2 - 14x + 36 = 0#

#1(x^2 - 14x + 49 - 49) = -36#

#1(x - 7)^2 - 49 = -36#

#1(x - 7)^2 = 13#

#(x - 7) = +-sqrt(13)#

#x = 7 +- sqrt(13)#

However, the #x = 7 - sqrt(13)# is extraneous, because it renders the initial equation undefined.

Hopefully this helps!