My question has to do with simplifying.

$\displaystyle e^(-9ln(x)) = 1/e^(9ln(x)) $

What does this simplify to?

[Math] 1/9x [/tex] ?

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- Mar 27th 2009, 01:18 PM #1

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- Mar 27th 2009, 01:24 PM #2

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- Mar 28th 2009, 09:56 AM #4

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A basic property of

**logs**leads to the conclusion that, for any base "b", the following holds:

. . . . .$\displaystyle b^{\log_b(m)}\, =\, m$

Applying a**log rule**inside that power to put that -9 back inside the log, we get a power of:

. . . . .$\displaystyle -9\ln(x)\, =\, \ln(x^{-9})\, =\, \ln\left(\frac{1}{x^9}\right)$

...by the**rule for negative exponents**.

Now apply the first rule above to simply the exponential expression.