How do you solve #log_2 (x-1) = log_4 (x+1)#?

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Answer 1 · 2016-01-12T23:28:14

#S = {0,3}#

Elevate four to both sides

#4^(log_2(x-1)) = 4^(log_4(x+1))#

Since #a^(log_a(b)) = b# we can simplify the RHS

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

But since #4 = 2^2#

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

So we have

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

Expanding

#x^2 - 2x + 1 = x + 1#
#x^2 - 2x - x + 1 - 1 = 0#
#x^2 - 3x = 0#

The roots are #x = 0# and #x = 3#