# simplification of.logarithms

• Nov 25th 2012, 03:56 PM
Tutu
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!
• Nov 25th 2012, 04:16 PM
skeeter
Re: simplification of.logarithms
$\displaystyle \log_4{y} = \frac{\log_2{y}}{\log_2{4}} = \frac{\log_2{y}}{2}$
• Nov 25th 2012, 04:25 PM
Tutu
Re: simplification of.logarithms
Thanks! But are my above simplifications correct?
• Nov 25th 2012, 04:26 PM
Soroban
Re: simplification of.logarithms
Hello, Tutu!

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

Quote:

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

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

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

Compare: .$\displaystyle \log_4(64) \,=\,3\:\text{ and }\:\log_2(64)\,=\6$
• Nov 25th 2012, 04:28 PM
skeeter
Re: simplification of.logarithms
Quote:

Originally Posted by Tutu
Thanks! But are my above simplifications correct?

I can't tell ... I don't know if you mean $\displaystyle \log_4{y}$ or $\displaystyle \log(4y)$
• Nov 26th 2012, 07:14 PM
Tutu
Re: simplification of.logarithms
4 is the base of the logarithm. Sorry for not being detailed and thank you so much!