Can someone please help me start on this:

http://82.221.28.13/webct/RelativeRe...s/image004.png

thanks

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- Apr 6th 2010, 09:19 AMbolsiusLogarithm question
Can someone please help me start on this:

http://82.221.28.13/webct/RelativeRe...s/image004.png

thanks

- Apr 6th 2010, 10:33 AMCaptainBlack
- Apr 6th 2010, 10:33 AMDeadstar
lol.

Can't see anything.

But while I'm here I may as well throw down some properties of logs in case they help...

$\displaystyle \log (ab) = \log(a) + \log(b)$.

$\displaystyle \log(a/b) = \log(a) - \log(b)$ as long as $\displaystyle b \neq 0$.

$\displaystyle \log(e^x) = x$ where [tex]e[/math[ is the exponential function. - Apr 6th 2010, 02:12 PMbolsius
- Apr 6th 2010, 02:18 PMCaptainBlack
- Apr 6th 2010, 02:19 PMDeadstar
- Apr 6th 2010, 02:43 PMbolsius
Yes thanks I got it...I wasnīt sure if I had to use the laws of logarithms both side of the equal sign (right and left)