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

1 Answer
AJ
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!

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