# Simplify (Lnx)^2,3,4,5 etc.

1. Mar 6, 2012

### GeekPioneer

1. The problem statement, all variables and given/known data
Im trying to simply (lnx)^n say (ln2)^3

2. Relevant equations

?

3. The attempt at a solution

Im lost on this one I can't find any information. All info I'm finding is dilute because of similar search terms.

2. Mar 6, 2012

### Hodgey8806

There's no way to simplify it in the form you have? Are you sure it isn't meant to be read as ln(x^n) Where the exponent is in the natural log? If so, the rules of logarithms tell us that we can move the n to the front and write it as n*ln(x). So, how would this look for say ln(2^3)?

3. Mar 6, 2012

### GeekPioneer

no its (ln2)^4 not ln2^4=4ln2 solving this I believe involves setting it equal to σ or some variable and then exponentiating using e. Although I could be completely wrong. I know this can be simplified and solve because my professor did it in class, I just can't find my notes from that day.

4. Mar 6, 2012

### Hodgey8806

Well, letting (ln2)^4 and trying to solve for say 2 here, we'd have to involve a 4th root:

Two solutions (for even roots):
ln2 = 4-root(y) and ln2 = -4-root(y)

So now, we can use e as:

2=e^(4-root(y)) and 2=e^(-4-root(y))=1/e^(4-root(y))

And of of course for odd roots it isn't necessary.

Does this look similar?

5. Mar 6, 2012

### GeekPioneer

I think I'm following you've if 4-root(y) means y1/4. Regardless that doesn't look right? Thanks for the interest though :). We should be able to get a concrete answer like 16ln2 or something if i remember correctly!

6. Mar 6, 2012

### Dick

No, ln(x)^n is just ln(x)^n. ln(x^n)=n*ln(x), but that's not what you are asking about.