anonymous
  • anonymous
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!! :)
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • 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
  • 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
  • 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
  • anonymous
Well, when one isn't? Idk- this is very confusing for me
anonymous
  • anonymous
So if log 2(64)=6, does that change it to 2^6=64? Do I have to solve for it?
SolomonZelman
  • SolomonZelman
yes 2\(^6\)=64 is the correct conversion
SolomonZelman
  • 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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