http://img692.imageshack.us/img692/9857/logd.jpg

Please can someone take me through it

Much appreciated (Happy)

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- Feb 10th 2010, 10:11 AM200001solving this log
http://img692.imageshack.us/img692/9857/logd.jpg

Please can someone take me through it

Much appreciated (Happy) - Feb 10th 2010, 10:21 AMearboth
You should use the following properties of the log-function:

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

$\displaystyle \log(a) - \log(b) = \log \left(\frac ab \right)$

$\displaystyle n \cdot \log(a) = \log(a^n)$

$\displaystyle \frac1n \cdot \log(a)=\log \left(\sqrt[n]{a} \right)$ - Feb 10th 2010, 10:26 AM200001
Hi

thanks

Im ok with the laws of logs but cant seem to get this one going for some reason - Feb 10th 2010, 10:35 AMe^(i*pi)
- Feb 10th 2010, 10:35 AMearboth
$\displaystyle {\color{blue}\frac14 \cdot \left(\log_5(5) - \log_5(3) \right) } + \frac13 \cdot \left(\log_5(2) + \log_5(7) \right)$

$\displaystyle {\color{blue}\log_5 \left(\sqrt[4]{ \frac53 } \right) }+ \frac13 \cdot \left(\log_5(2) + \log_5(7) \right)$

I've transformed the first summand according the laws of logarithms. I'll leave the second summand for you. Afterwards combine both summands to get one term.