AonZ
  • AonZ
Last question!!! log 1/5 (25)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AonZ
  • AonZ
check the identity by evaluating...
AonZ
  • AonZ
\[\huge \log_{\frac{ 1 }{ 5 }}25 \]
hartnn
  • hartnn
use the same property i gave for last Q

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More answers

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

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