# How do you condense 1/3(log_8y+2log_8(y+4))-log_8(y-1)?

Jan 25, 2017

${\log}_{8} \left(\frac{\sqrt[3]{y {\left(y + 4\right)}^{2}}}{y - 1}\right)$

#### Explanation:

Since $k \log a = \log {a}^{k}$, the expression is equivalent to:

$\frac{1}{3} \left({\log}_{8} y + {\log}_{8} {\left(y + 4\right)}^{2}\right) - {\log}_{8} \left(y - 1\right)$

Since $\log a + \log b = \log \left(a b\right)$, you get:

$\frac{1}{3} {\log}_{8} y {\left(y + 4\right)}^{2} - {\log}_{8} \left(y - 1\right)$

$= {\log}_{8} {\left(y {\left(y + 4\right)}^{2}\right)}^{\frac{1}{3}} - {\log}_{8} \left(y - 1\right)$

Since $\log a - \log b = \log \left(\frac{a}{b}\right)$, you get

${\log}_{8} \left({\left(y {\left(y + 4\right)}^{2}\right)}^{\frac{1}{3}} / \left(y - 1\right)\right)$

that can be written as:

${\log}_{8} \left(\frac{\sqrt[3]{y {\left(y + 4\right)}^{2}}}{y - 1}\right)$