How do you solve #log_3 (2x-4) - log_3 (x-1)=log_3 (x-2)#?

1 Answer
Gió
Dec 21, 2015

I found:
#x=3#

Explanation:

We can use the property of logs:
#logx-logy=log(x/y)#
And write:
#log_3[(2x-4)/(x-1)]=log_3(x-2)#
If the logs are equal the arguments must be as well, so:
#(2x-4)/(x-1)=x-2#
#2x-4=(x-1)(x-2)#
#2x-4=x^2-2x-x+2#
#x^2-5x+6=0#
Use the Quadratic Formula:
#x_(1,2)=(5+-sqrt(25-24))/2=#
Two solutions:
#x_1=3# YES
#×_2=2# NO

Related questions
See all questions in Logarithmic Models
Impact of this question
3687 views around the world
You can reuse this answer
Creative Commons License