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