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anonymous
 one year ago
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!! :)
anonymous
 one year ago
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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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1There 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
 one year ago
Best ResponseYou've already chosen the best response.1For 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
 one year ago
Best ResponseYou've already chosen the best response.0That was very helpful, thank you but I don't see how that helps when they're not in logarithmic form..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, when one isn't? Idk this is very confusing for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if log 2(64)=6, does that change it to 2^6=64? Do I have to solve for it?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes 2\(^6\)=64 is the correct conversion

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and 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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