Question on Logs

**davyboy123** · November 29th 2009, 08:30 AM

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)

**Jhevon** · November 29th 2009, 08:33 AM

Originally Posted by davyboy123

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)

Note that, by the property $n \log X = \log X^n$ , you have $\log x^{2/3} + \log x + \log x^{-9/4}$ . Now use the property $\log X + \log Y = \log XY$

**skeeter** · November 29th 2009, 08:38 AM

Originally Posted by davyboy123

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)

I agree with you ...

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

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

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

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

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

$\log\left(x^{-\frac{7}{12}}\right)<br />$

... books have been wrong before.

**mrmohamed** · December 1st 2009, 03:10 PM

Originally Posted by davyboy123

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

Originally Posted by davyboy123

1/3log(x^2) + log x - 3/4 log (x^3)
In the book it says log (x^1/12)

Hi all
another methode

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