Thread: e, ln(x)

My question has to do with simplifying.

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

What does this simplify to?

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

Originally Posted by KyleC

My question has to do with simplifying.

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

What does this simplify to?

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

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

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

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

Combining and cancelling:

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

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

Ahh, Thank you very much.

Originally Posted by KyleC

$\displaystyle 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:

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

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.