How to Simplify (Lnx)^n for Homework?

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Homework Help Overview

The discussion revolves around simplifying the expression (lnx)^n, specifically with an example of (ln2)^3. Participants are exploring the nature of logarithmic expressions and their simplifications.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question whether (lnx)^n could be interpreted as ln(x^n) instead, which would allow for simplification using logarithmic properties. Others express uncertainty about the simplification process and mention the need for a variable substitution or exponentiation.

Discussion Status

The discussion is ongoing, with participants sharing different interpretations and approaches. There is no explicit consensus on the simplification method, but various lines of reasoning are being explored.

Contextual Notes

Participants note the challenge of finding clear information on the topic and mention a lack of access to class notes that might clarify the professor's explanation. There is also a reference to homework constraints that may affect the discussion.

GeekPioneer
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Homework Statement


Im trying to simply (lnx)^n say (ln2)^3

Homework Equations



?


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.

Please, and thanks!
 
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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)?
 
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.
 
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?
 
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!
 
GeekPioneer said:
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!

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

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