Last question!!! log 1/5 (25)

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Last question!!! log 1/5 (25)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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check the identity by evaluating...
\[\huge \log_{\frac{ 1 }{ 5 }}25 \]
use the same property i gave for last Q

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\(\frac{1}{5} = 5^{-1}\). What would you multiply to that to get \(5^2\)?
|dw:1360930788428:dw|
yes, simplify numerator and denominator separately.
\[\large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{5} 25 }{ \log_{5}\frac{ 1 }{ 5}} = \frac{ \log_{5} 5^2 }{ \log_{5} 5^{-1}}\]
Why so serious?
Just figure out what you multiply to \(5^{-1}\) to get \(5^2.\)
ahh @agent0smith explained it pretty good :)
@agent0smith How did you get the base to be 5? Like all of them?
Not sure what you mean... I just used base 5 because it makes it easy to solve.
@AonZ Change of bases.
you can choose any base for using that property,e,5,10,... thats why its unspecified in the property
so if you use change of base rule you can choose ANY base?
Yes.
wow i reckon i understand logs a lot better now :P
^ correct @ParthKohli shouldn't it be "figure out what power you need to raise 5^-1 to, to get 25" as opposed to "Just figure out what you multiply to 5^−1 to get 5^2"?
Yes, I am sorry.
But you get the point.
Ah okay, cos i was wondering how this helps:\[5^{-1} \times 5^3 = 25\]compared to this \[(5^{-1})^{-2}=25\]
:-)
@AonZ you could also do it this way (but it makes it a bit more difficult): \[ \large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{25} 25 }{ \log_{25}\frac{ 1 }{ 5}} = \frac{1 }{ \log_{25} 25^{-0.5}}\]
or simply \(\huge \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{} 5 }{ \log_{}\frac{ 1 }{ 5}} \)
Much better up there.
\(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}\)
^ that works fine, but requires a calculator.
Nope.
\(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}\)
calcy for what ?
\[\dfrac{\log _5 5^2}{\log _5 5^{-1}} = \]
I'm assuming @hartnn is using base 10...
\[\dfrac{2}{-1}\]
does not matter what base i use, by default its 10
You can use change of base too.
\[ \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}} \] I mean, how are you solving this w/o changing base again, or using a calculator?
\(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}=\dfrac{2 \log 5}{-1 \log 5}=-2\)
Oh you're simplifying to \[ \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}} = \frac{ 2 \log5 }{ -\log5 }\]

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