## 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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1. SolomonZelman

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

2. SolomonZelman

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

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. SolomonZelman

yes 2$$^6$$=64 is the correct conversion

7. 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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