How do you solve for x in log_2(x-3)=2-log_2(x-6)?

1 Answer
Apr 8, 2018

x=2,7

Explanation:

Given log_2 (x-3)=2-log_2 (x-6)
or, log_2 (x-3)+log (x-6)=2
or, log_2 [(x-3)(x-6)]=2
or, log_2 (x^2-9x+18)=2
or, log_2 (x^2-9x+18)=log_2 (4)
By one to one property,
(x^2-9x+18)=4
or, x^2-9x+18-4=0
or, x^2-9x+14=0
or, x^2-7x-2x+14=0
or, (x-2)(x-7)=0
x=2,7
Thank you!