How do you solve #log_2 (x - 1) – log_2 (x – 4) = log_2 4#?

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Answer 1 · 2015-12-01T11:19:30

#x=5#

We have that #log(a)-log(b)=log(a/b)#, so

#log_2(x-1)-log_2(x-4) = log_2((x-1)/(x-4))#

The equation has thus become

#log_2((x-1)/(x-4))=log_2(4)#

This means that

#(x-1)/(x-4) = 4#

If #x \ne 4#, multiply both sides by #x-4#, obtaining

#x-1 = 4x-16#, and solve as usual

#-3x = -15 \to x=5#

Verify the answer:

If #x=5#, the expression becomes

#log_2(5-1) - log_2(5-4) = log_2(4)#

#log_2(4) - log_2(1) = log_2(4)#

which is true, since #log_a(1)=0# for every #a#.