# quick help with logs

• Dec 18th 2007, 02:27 AM
incyt3s
quick help with logs
solve:
$\log_{7} \left ( {\frac{1}
{\sqrt{7}}} \right )$

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

or

$\log_{7}1 - \log_{7}\sqrt{7} = x$
• Dec 18th 2007, 03:58 AM
janvdl
Quote:

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

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

or

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

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

• Dec 18th 2007, 05:29 AM
incyt3s
yess

thanks
• Dec 19th 2007, 12:26 AM
Aryth
Quote:

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

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

or

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

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

$\log_{7}1 = 0$

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

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

$\log_{7}7 = 1$

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