How do you solve #log _2 (x-1) = log _4 (x+1)#?

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Answer 1 · 2015-12-06T11:00:14

#x=3#

First of all, you need to change the base of the logarithm: since #4=2^2#, you have that

#log_4(x) = log_2(x)/log_2(4) = log_2(x)/2#

So, the equation becomes

#log_2(x-1) = log_2(x+1)/2#

Which is equivalent to

#2log_2(x-1) = log_2(x+1)#

Since #alog(x)=log(x^a)#, the equation becomes

#log_2((x-1)^2) = log_2(x+1)#

Which is true if and only if #(x-1)^2=x+1#, which means

#x^2-2x+1=x+1 \to x^2-3x=0#.

This equation is solved by #x=0# and #x=3#, but #x=0# is not acceptable, because it would lead to

#log_2(-1)=log_4(1)#, and we can't compute logarithms of negative numbers.

As for #x=3#, we can see that it's ok since it leads to

#log_2(2) = log_4(4)#, which is true because both members are equal to #1#.