anonymous
  • anonymous
What is the value of
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\log_{81} 3\]
anonymous
  • anonymous
anonymous
  • anonymous

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DecentNabeel
  • DecentNabeel
\[\log _{81}\left(3\right)=\frac{1}{4}\quad \left(\mathrm{Decimal:\quad }\:0.25\right)\]
anonymous
  • anonymous
DecentNabeel
  • DecentNabeel
Happy:) @EllenJaz17
anonymous
  • anonymous
how do you value that though? I have to show my work
anonymous
  • anonymous
DecentNabeel
  • DecentNabeel
\[\mathrm{Rewrite\:}3\mathrm{\:\in\:power-base\:form:}\quad 3=81^{\frac{1}{4}}\] \[=\log _{81}\left(81^{\frac{1}{4}}\right)\] \[\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\] \[\log _{81}\left(81^{\frac{1}{4}}\right)=\frac{1}{4}\log _{81}\left(81\right)\] \[=\log _{81}\left(81\right)\frac{1}{4}\] \[\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(a\right)=1\] \[\log _{81}\left(81\right)=1\] =1/4
DecentNabeel
  • DecentNabeel
anonymous
  • anonymous
\[\log_{81}(3)=x\iff 81^x=3\] so what you are trying to do is figure out how to write 3 as a power of 81 since \(\sqrt[4]{81}=3\) that tells you \[81^{\frac{1}{4}}=3\] and you have the power you are looking for
anonymous
  • anonymous
so it's not 1/4 its 3?
anonymous
  • anonymous
another easier example \[\log_7(49)=2\] because \(7^2=49\)
anonymous
  • anonymous
no the answer is \(\frac{1}{4}\) let me write it again
anonymous
  • anonymous
ok sounds good
anonymous
  • anonymous
\[\huge \log_{81}(\color{red}x)=3\iff 81^{\color{red}x}=3\] you are trying to find \(\color{red}x\)
anonymous
  • anonymous
you are looking for the power that you would raise 81 to to give you 3
anonymous
  • anonymous
\[\log_2(8)=3\iff 2^3=8\] they say the same thing
anonymous
  • anonymous
you should be able to switch back and forth quickly
anonymous
  • anonymous
the only tricky part of this one was recognizing that 3 is the fourth root of 81
anonymous
  • anonymous
.sorry catching up was getting something to drink
anonymous
  • anonymous
beer i hope
anonymous
  • anonymous
haha no just water
anonymous
  • anonymous
too bad
anonymous
  • anonymous
I prefer straight tequila, gets the job done quick lol
anonymous
  • anonymous
no beer chaser? dang!
anonymous
  • anonymous
Haha nope!
anonymous
  • anonymous
oops do you get that \(2^5=32\) says the same thing as \(\log_2(32)=5\) ?
anonymous
  • anonymous
how would i apply that to find this value in my question?
anonymous
  • anonymous
so if you were trying to solve \[\log_2(32)=\color{red}x\] you would be trying to solve \[2^{\color{red}x}=32\]
anonymous
  • anonymous
you have \[\log_{81}(3)=\color{red}x\] you have to solve \[81^{\color{red}x}=3\]
anonymous
  • anonymous
since \(\sqrt[4]{81}=x\) that is the same think as saying \(81^{\frac{1}{4}}=3\)
anonymous
  • anonymous
ooops i means "since \(\sqrt[4]{81}=3\)
anonymous
  • anonymous
now you have the power you are looking for , that power is \(\frac{1}{4}\) therefore since \[81^{\frac{1}{4}}=3\] it is the same as saying \[\log_{81}(3)=\frac{1}{4}\]
anonymous
  • anonymous
Ok i kinda get it lol
anonymous
  • anonymous
yeah it is confusing at first a simpler example is it clear that \(10^4=10,000\)?

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