How do you solve ${log}_{2} x + {log}_{2} (x - 1) = 2$ $log_ 2x + log_2(x-1)=2$ ?

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Answer 1 · 2016-05-06T17:45:14

$x = \frac{1 + \sqrt{17}}{2}$ $x=(1+sqrt17)/2$

${log}_{2} x + {log}_{2} (x - 1) = 2$ $log_2x+log_2(x-1)=2$ (note $x$ $x$ cannot be less than $1$ $1$ )

or ${log}_{2} x (x - 1) = {log}_{2} (4)$ $log_2x(x-1)=log_2(4)$

or $x (x - 1) = 4$ $x(x-1)=4$

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

Using quadratic formula we get

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \cdot 1 \cdot (- 4)}}{2}$ $x=(-(-1)+-sqrt((-1)^2-4*1*(-4)))/(2$

or $x = \frac{1 \pm \sqrt{17}}{2}$ $x=(1+-sqrt17)/2$

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

$x = \frac{1 + \sqrt{17}}{2}$ $x=(1+sqrt17)/2$

How do you solve log2x+log2(x−1)=2log_ 2x + log_2(x-1)=2?