# Math Help - simplification of.logarithms

1. ## simplification of.logarithms

I was wondering of thsee simplifications are possible, log4(y) =log2^2(y) =log2(y)^2 OR logy since.log2^2 is 1. Thank you.so.much!

2. ## Re: simplification of.logarithms

$\log_4{y} = \frac{\log_2{y}}{\log_2{4}} = \frac{\log_2{y}}{2}$

3. ## Re: simplification of.logarithms

Thanks! But are my above simplifications correct?

4. ## Re: simplification of.logarithms

Hello, Tutu!

Very imaginative thinking . . . sadly, it doesn't work.

I was wondering of thsee simplifications are possible:
. . $\log_4(y) \:=\:\log_{2^2}(y) \:=\:\log_2(y^2)\:\text{ or }\:\log_2(y)\:\text{ since }\log_22 \,=\,1.$

That last part is false: . $\log_2(y^2)$ does not equal $\log_2(y).$

And the first part is false; we cannot move the "2" like that.

Compare: . $\log_4(64) \,=\,3\:\text{ and }\:\log_2(64)\,=\6$

5. ## Re: simplification of.logarithms

Originally Posted by Tutu
Thanks! But are my above simplifications correct?
I can't tell ... I don't know if you mean $\log_4{y}$ or $\log(4y)$

6. ## Re: simplification of.logarithms

4 is the base of the logarithm. Sorry for not being detailed and thank you so much!