# Thread: Exponential Rules

1. ## Exponential Rules

Good afternoon everybody...

I need some help on the following - I'm sure it's simple, but I've totallk forgot how to do it...

Say I wanted to take the exponential of the following (where a and b are constants).

ln(a*x)^b

which, is the same as

b*ln(ax)^b

what would be the result?

I really appreciate your help on this - I have searched the web - I'm totally blank!

2. ## Re: Exponential Rules

$\displaystyle \ln(ax)^b = b\ln(ax) = b(\ln{a} + \ln{x}) = b\ln{a} + b\ln{x}$

note that this relationship also works right to left ...

3. ## Re: Exponential Rules

Many thanks Skeeter, that was a great help - another question, if I may - along the same ilk...

I have this:

$\displaystyle \ln(ax+b)^k+c$

Now, taking an exponential, do I get...

$\displaystyle \exp(\ln(ax+b)^k+c) = (ax+b)^k+\exp(c)$

I can't find any info on this rule anywhere

Many thanks for your helo so far - reps for you :-)

4. ## Re: Exponential Rules

Originally Posted by MaverickUK82
Many thanks Skeeter, that was a great help - another question, if I may - along the same ilk...

I have this:

$\displaystyle \ln(ax+b)^k+c$

Now, taking an exponential, do I get...

$\displaystyle \exp(\ln(ax+b)^k+c) = (ax+b)^k+\exp(c)$

I can't find any info on this rule anywhere

Many thanks for your helo so far - reps for you :-)
Not quite you rule you are looking for is

$\displaystyle e^{x+y}=e^{x}e^{y}$

So you should end up with

$\displaystyle e^{\ln(ax+b)^k+c}=e^{\ln(ax+b)^k}e^{c}=(ax+b)^ke^{ c}$