I really don't understand logarithms... I have two questions and I would really appreciate if someone could explain them to me and help me!!
1. Rewrite log 2(64)=6 as an exponential.
(Read log base 2 of 64)
2. Rewrite x=a^y as an logarithm.
Thank you in advance!! :)

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- anonymous

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- SolomonZelman

There is a rule:
\(\Large\color{black}{ \displaystyle \log_{\color{red}{\rm A}}\left(\color{green}{\rm B}\right)=\color{blue}{\rm C}{~~~~~~} _{\Huge ~~\Longrightarrow}^{\rm converts~to }{~~~~~~} \color{red}{\rm A}^\color{blue}{\rm C}=\color{green}{\rm B}}\)

- SolomonZelman

For example,
\(\large\color{black}{ \displaystyle \log_3x=4 }\)
would convert to
\(\large\color{black}{ \displaystyle 3^4=x }\)
and then
\(\large\color{black}{ \displaystyle x=81 }\)

- anonymous

That was very helpful, thank you- but I don't see how that helps when they're not in logarithmic form..

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- anonymous

Well, when one isn't? Idk- this is very confusing for me

- anonymous

So if log 2(64)=6, does that change it to 2^6=64? Do I have to solve for it?

- SolomonZelman

yes 2\(^6\)=64 is the correct conversion

- SolomonZelman

and you don't need to solve for anything here:)
2\(^6\) = (2\(^3\))\(^2\) = 8\(^2\) = 64,
so you know it is true...

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