How do you solve log_3 (2x-4) - log_3 (x-1)=log_3 (x-2)log3(2x−4)−log3(x−1)=log3(x−2)?

1 Answer
Gió
Dec 21, 2015

I found:
x=3x=3

Explanation:

We can use the property of logs:
logx-logy=log(x/y)logx−logy=log(xy)
And write:
log_3[(2x-4)/(x-1)]=log_3(x-2)log3[2x−4x−1]=log3(x−2)
If the logs are equal the arguments must be as well, so:
(2x-4)/(x-1)=x-22x−4x−1=x−2
2x-4=(x-1)(x-2)2x−4=(x−1)(x−2)
2x-4=x^2-2x-x+22x−4=x2−2x−x+2
x^2-5x+6=0x2−5x+6=0
Use the Quadratic Formula:
x_(1,2)=(5+-sqrt(25-24))/2=x1,2=5±√25−242=
Two solutions:
x_1=3x1=3 YES
×_2=2×2=2 NO

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