# How do you solve 2log_5(x-2)=log_5 36?

Nov 3, 2016

#### Explanation:

Given:

$2 {\log}_{5} \left(x - 2\right) = {\log}_{5} \left(36\right)$

Divide both sides by 2:

${\log}_{5} \left(x - 2\right) = \left(\frac{1}{2}\right) {\log}_{5} \left(36\right)$

Use the property $\left(c\right) {\log}_{b} \left(a\right) = {\log}_{b} \left({a}^{c}\right)$

${\log}_{5} \left(x - 2\right) = {\log}_{5} \left({36}^{\frac{1}{2}}\right)$

Replace ${36}^{\frac{1}{2}}$ with 6:*

${\log}_{5} \left(x - 2\right) = {\log}_{5} \left(6\right)$

*Please notice that, technically, ${36}^{\frac{1}{2}} = \pm 6$ but the negative value would violate the domain of the logarithm.

Make the logarithms disappear by writing both sides as exponents to base.

${5}^{{\log}_{5} \left(x - 2\right)} = {5}^{{\log}_{5} \left(6\right)}$

$x - 2 = 6$

$x = 8$

Nov 3, 2016

$x = 8$

#### Explanation:

$2 {\log}_{5} \left(x - 2\right) = {\log}_{5} 36 \text{ } \leftarrow$ use the log power law

${\log}_{5} {\left(x - 2\right)}^{2} = {\log}_{5} 36$

Note: If $\log A = \log B \Leftrightarrow A = B$

$\therefore {\left(x - 2\right)}^{2} = 36 \text{ } \leftarrow$ find the square root

$x - 2 = \sqrt{36} = \pm 6 \text{ } \leftarrow$ reject -6

$x - 2 = 6$

$x = 8$