Log question. Is there a rule that can solve this? \[\huge \log_{\frac{ 1 }{ a }} \frac{ 1 }{ n }\]

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Log question. Is there a rule that can solve this? \[\huge \log_{\frac{ 1 }{ a }} \frac{ 1 }{ n }\]

Mathematics
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@agent0smith can you help with this?
Yeah i'm not sure what you want to do with it... you can change it into a couple of different forms?
There is a formula for changing the base of the logarithm.

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well il type full equation prove log a(x) = -log 1/2 (x)
use : \(\huge \log_ab=\dfrac{\log b}{\log a}\)
u sure, this is the Q ? prove log a(x) = -log 1/2 (x)
wait my bad LOL log a(x) = -log 1/a (x)
right, use the property i gave on RIGHT side.
|dw:1360928528477:dw|
|dw:1360928591721:dw| these cancel right?
simplify both numerator and denominator. and - log x = log(1/x) was unnecessary and no, not directly.
how else do i do it then??
|dw:1360928748794:dw|
oh so the negatives cancel?
yesh, now negatives can cancel.
again apply that property on what you have...
\(\huge \dfrac{\log b}{\log a}=\log_ab\)
yes i can see how the equal each other now :)
\[\log_{a} x = -\log_{\frac{ 1 }{ a }} x\] Use the log rule on the right side \[-\log_{\frac{ 1 }{ a }} x =- \frac{ \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } =- \frac{ \log_{a} x }{ \log_{a} a ^{-1}}=- \frac{ \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}\]
moved the negatives, makes it easier to follow. \[-\log_{\frac{ 1 }{ a }} x = \frac{ - \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } = \frac{- \log_{a} x }{ \log_{a} a ^{-1}}= \frac{- \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}\]
ahh ok thanks to all of you :)

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