What is the value of

- anonymous

What is the value of

- chestercat

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- anonymous

\[\log_{81} 3\]

- anonymous

- anonymous

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## More answers

- DecentNabeel

\[\log _{81}\left(3\right)=\frac{1}{4}\quad \left(\mathrm{Decimal:\quad }\:0.25\right)\]

- anonymous

- DecentNabeel

Happy:) @EllenJaz17

- anonymous

how do you value that though? I have to show my work

- anonymous

- 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

ok @EllenJaz17

- 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

so it's not 1/4 its 3?

- anonymous

another easier example
\[\log_7(49)=2\] because \(7^2=49\)

- anonymous

no the answer is \(\frac{1}{4}\) let me write it again

- anonymous

ok sounds good

- anonymous

\[\huge \log_{81}(\color{red}x)=3\iff 81^{\color{red}x}=3\] you are trying to find \(\color{red}x\)

- anonymous

you are looking for the power that you would raise 81 to to give you 3

- anonymous

\[\log_2(8)=3\iff 2^3=8\] they say the same thing

- anonymous

you should be able to switch back and forth quickly

- anonymous

the only tricky part of this one was recognizing that 3 is the fourth root of 81

- anonymous

.sorry catching up was getting something to drink

- anonymous

beer i hope

- anonymous

haha no just water

- anonymous

too bad

- anonymous

I prefer straight tequila, gets the job done quick lol

- anonymous

no beer chaser? dang!

- anonymous

Haha nope!

- anonymous

oops
do you get that \(2^5=32\) says the same thing as \(\log_2(32)=5\) ?

- anonymous

how would i apply that to find this value in my question?

- 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

you have
\[\log_{81}(3)=\color{red}x\] you have to solve
\[81^{\color{red}x}=3\]

- anonymous

since \(\sqrt[4]{81}=x\) that is the same think as saying \(81^{\frac{1}{4}}=3\)

- anonymous

ooops i means "since \(\sqrt[4]{81}=3\)

- 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

Ok i kinda get it lol

- anonymous

yeah it is confusing at first
a simpler example
is it clear that \(10^4=10,000\)?

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