# How do you solve log_2(x-6)+log_2(x- 4)=log_2x?

Apr 16, 2018

$\implies x = 8 , 3$

#### Explanation:

${\log}_{2} \left(x - 6\right) + {\log}_{2} \left(x - 4\right) = {\log}_{2} x$

$\implies {\log}_{2} \left[\left(x - 6\right) \left(x - 4\right)\right] = {\log}_{2} x$

$\implies \left(x - 6\right) \left(x - 4\right) = x$

$\implies {x}^{2} - 10 x + 24 - x = 0$

$\implies {x}^{2} - 11 x + 24 = 0$

$\implies {x}^{2} - 8 x - 3 x + 24 = 0$

$\implies x \left(x - 8\right) - 3 \left(x - 8\right) = 0$

$\implies \left(x - 8\right) \left(x - 3\right) = 0$

$\implies x = 8 , 3$

Apr 16, 2018

$\textcolor{b l u e}{x = 8}$

#### Explanation:

By the laws of logarithms:

log_a(b)+log)a(c)=log_a(bc)

${\log}_{2} \left(x - 6\right) + {\log}_{2} \left(x - 4\right) = {\log}_{2} \left(\left(x - 6\right) \left(x - 4\right)\right) = {\log}_{2} \left(x\right)$

If:

${\log}_{2} \left(\left(x - 6\right) \left(x - 4\right)\right) = {\log}_{2} \left(x\right)$

Then:

(x-6)(x-4)=x#

Expanding and simplifying:

${x}^{2} - 10 x + 24 = x$

${x}^{2} - 11 x + 24 = 0$

Factor:

$\left(x - 3\right) \left(x - 8\right) = 0 \implies x = 3 \mathmr{and} x = 8$

For $x = 3$

${\log}_{2} \left(\left(3\right) - 6\right) = {\log}_{2} \left(- 3\right)$

This is not defined for real numbers:

$\textcolor{b l u e}{x = 8}$ is the only solution: