# Thread: quick help with logs

1. ## quick help with logs

solve:
$\displaystyle \log_{7} \left ( {\frac{1} {\sqrt{7}}} \right )$

$\displaystyle 7^x = \frac{1}{\sqrt{7}}$

or

$\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$

2. Originally Posted by incyt3s
solve:
$\displaystyle \log_{7} \left ( {\frac{1} {\sqrt{7}}} \right )$

$\displaystyle 7^x = \frac{1}{\sqrt{7}}$

or

$\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$
$\displaystyle \log_{7} \left( \frac{ 1 }{ \sqrt{7} } \right) = \log_{7} \left( 7^{- \frac{1}{2} } \right)$

3. yess

thanks

4. Originally Posted by incyt3s
solve:
$\displaystyle \log_{7} \left ( {\frac{1} {\sqrt{7}}} \right )$

$\displaystyle 7^x = \frac{1}{\sqrt{7}}$

or

$\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$

$\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = \log_{7}1 - \log_{7}\sqrt{7}$

$\displaystyle \log_{7}1 = 0$

$\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = - \log_{7}\sqrt{7}$

$\displaystyle \log_{7}\sqrt{7} = \log_{7}7^\frac{1}{2} = \frac{1}{2}\log_{7}7$

$\displaystyle \log_{7}7 = 1$

$\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = - \frac{1}{2}$