e, ln(x)

**KyleC** · Mar 27th 2009, 01:18 PM

My question has to do with simplifying.

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

What does this simplify to?

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

**e^(i*pi)** · Mar 27th 2009, 01:24 PM

Originally Posted by KyleC

My question has to do with simplifying.

$e^{(-9ln(x))} = 1/e^{(9ln(x))}$

What does this simplify to?

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

$kln(a) = ln(a^k)$

$e^{-9ln(x)} = e^{ln(x^{-9})}$

$\frac{1}{e^{9ln(x)}} = \frac{1}{e^{ln({x^9})} =$

Combining and cancelling:

$x^{-9} = \frac{1}{x^9}$

$x^9$ to give $1=1$

**KyleC** · Mar 27th 2009, 01:27 PM

Ahh, Thank you very much.

**stapel** · Mar 28th 2009, 09:56 AM

Originally Posted by KyleC

$e^{-9ln(x)} = 1/e^{9ln(x)}$

What does this simplify to?

A basic property of logs leads to the conclusion that, for any base "b", the following holds:

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

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

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

...by the rule for negative exponents.

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

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