Just to be sure i think i did this right i had to simplify

$\displaystyle \ln \frac{x^2 e}{y^3 e^x}$

$\displaystyle = \ln x^2 + \ln e - \ln y^3 + \ln e^x$

$\displaystyle = 2 \ln x + 1 - (3 \ln y + x)$

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- Feb 20th 2010, 09:59 AMx5pyd3rxOne more Question on rewriting logs
Just to be sure i think i did this right i had to simplify

$\displaystyle \ln \frac{x^2 e}{y^3 e^x}$

$\displaystyle = \ln x^2 + \ln e - \ln y^3 + \ln e^x$

$\displaystyle = 2 \ln x + 1 - (3 \ln y + x)$ - Feb 20th 2010, 10:31 AMe^(i*pi)
Yep, no problem there