$\displaystyle ln (e^{x^2-y^2})$

Anyone able to help me?

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- Jul 4th 2010, 02:08 PMorkdorkNatural Logs help
$\displaystyle ln (e^{x^2-y^2})$

Anyone able to help me? - Jul 4th 2010, 02:11 PMpickslides
Follow this.

$\displaystyle \log_ab^c = c\log_ab$ - Jul 4th 2010, 02:11 PMAlso sprach Zarathustra
Use this formula:

$\displaystyle ln (a)^b=b \cdot ln(a)$ - Jul 6th 2010, 01:09 AMHallsofIvy
Another way to do basically what pickslides and Also sprach Zarathustra suggest is to use the fact that ln(x) and $\displaystyle e^x$ are

**inverse**functions: $\displaystyle ln(e^x)= x$. - Jul 6th 2010, 07:12 AMmr fantastic