How do you solve #log_ 2x + log_2(x-1)=2#?

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Answer 1 · 2016-05-06T17:45:14

#x=(1+sqrt17)/2#

#log_2x+log_2(x-1)=2# (note #x# cannot be less than #1#)

or #log_2x(x-1)=log_2(4)#

or #x(x-1)=4#

or #x^2-x-4=0#

Using quadratic formula we get

#x=(-(-1)+-sqrt((-1)^2-4*1*(-4)))/(2#

or #x=(1+-sqrt17)/2#

As #x# cannot be less than #1#,

#x=(1+sqrt17)/2#