How do you solve for x in log_2(x-3)=2-log_2(x-6)?

Apr 8, 2018

$x = 2 , 7$

Explanation:

Given ${\log}_{2} \left(x - 3\right) = 2 - {\log}_{2} \left(x - 6\right)$
$\mathmr{and} , {\log}_{2} \left(x - 3\right) + \log \left(x - 6\right) = 2$
$\mathmr{and} , {\log}_{2} \left[\left(x - 3\right) \left(x - 6\right)\right] = 2$
$\mathmr{and} , {\log}_{2} \left({x}^{2} - 9 x + 18\right) = 2$
$\mathmr{and} , {\log}_{2} \left({x}^{2} - 9 x + 18\right) = {\log}_{2} \left(4\right)$
By one to one property,
$\left({x}^{2} - 9 x + 18\right) = 4$
$\mathmr{and} , {x}^{2} - 9 x + 18 - 4 = 0$
$\mathmr{and} , {x}^{2} - 9 x + 14 = 0$
$\mathmr{and} , {x}^{2} - 7 x - 2 x + 14 = 0$
$\mathmr{and} , \left(x - 2\right) \left(x - 7\right) = 0$
$x = 2 , 7$
Thank you!