quick help with logs

**incyt3s** · December 18th 2007, 02:27 AM

solve:
$\log_{7} \left ( {\frac{1}<br /> {\sqrt{7}}} \right )$

how would you go about this w/o a calculator

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

or

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

**janvdl** · December 18th 2007, 03:58 AM

Originally Posted by incyt3s

solve:
$\log_{7} \left ( {\frac{1}<br /> {\sqrt{7}}} \right )$

how would you go about this w/o a calculator

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

or

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

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

Does this help you?

**incyt3s** · December 18th 2007, 05:29 AM

yess

thanks

**Aryth** · December 19th 2007, 12:26 AM

Originally Posted by incyt3s

solve:
$\log_{7} \left ( {\frac{1}<br /> {\sqrt{7}}} \right )$

how would you go about this w/o a calculator

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

or

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

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

$\log_{7}1 = 0$

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

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

$\log_{7}7 = 1$

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

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