When I try and solve the following I get the answer as log (x^-7/12)

1/3log(x^2) + log x - 3/4 log (x^3)

In the book it says log (x^1/12)

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- Nov 29th 2009, 08:30 AMdavyboy123Question on Logs
When I try and solve the following I get the answer as log (x^-7/12)

1/3log(x^2) + log x - 3/4 log (x^3)

In the book it says log (x^1/12) - Nov 29th 2009, 08:33 AMJhevon
- Nov 29th 2009, 08:38 AMskeeter
I agree with you ...

$\displaystyle \frac{1}{3}\log(x^2) + \log{x} - \frac{3}{4}\log(x^3)$

$\displaystyle \log(x^{\frac{2}{3}}) + \log{x} - \log(x^{\frac{9}{4}})$

$\displaystyle \log\left(\frac{x^{\frac{2}{3}} \cdot x}{x^{\frac{9}{4}}}\right)$

$\displaystyle \log\left(\frac{x^{\frac{5}{3}}}{x^{\frac{9}{4}}}\ right)$

$\displaystyle \log\left(\frac{x^{\frac{20}{12}}}{x^{\frac{27}{12 }}}\right)$

$\displaystyle \log\left(x^{-\frac{7}{12}}\right)

$

... books have been wrong before. - Dec 1st 2009, 03:10 PMmrmohamed