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

May 6, 2016

$x = \frac{1 + \sqrt{17}}{2}$

#### Explanation:

${\log}_{2} x + {\log}_{2} \left(x - 1\right) = 2$ (note $x$ cannot be less than $1$)

or ${\log}_{2} x \left(x - 1\right) = {\log}_{2} \left(4\right)$

or $x \left(x - 1\right) = 4$

or ${x}^{2} - x - 4 = 0$

x=(-(-1)+-sqrt((-1)^2-4*1*(-4)))/(2
or $x = \frac{1 \pm \sqrt{17}}{2}$
As $x$ cannot be less than $1$,
$x = \frac{1 + \sqrt{17}}{2}$