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 $1$ ,

$x=(1+sqrt17)/2$

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