Log. functions - comparing solutions, which is best?

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Homework Help Overview

The discussion revolves around evaluating logarithmic expressions, specifically comparing two different interpretations of the logarithm of a power. Participants are examining the validity of two proposed solutions to a logarithmic equation involving the expression log_x(x^n).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the correctness of two different solutions to the logarithmic expression, with one solution suggesting that log_x(x^n) simplifies to n, while the other suggests it simplifies to 1. There is an exploration of the notation used and its implications on the interpretation of the logarithm.

Discussion Status

Some participants have provided guidance on the correct interpretation of the logarithmic expressions, noting the importance of understanding the placement of parentheses in the notation. Multiple interpretations are being explored, and there is a focus on clarifying the assumptions behind each solution.

Contextual Notes

Participants are discussing potential errors in notation and the implications of these errors on the solutions presented. There is an emphasis on the need for clarity in mathematical expressions to avoid confusion.

Tyrannosaurus_
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Homework Statement


I'm not sure what my error is. Both solutions cannot be true.

Homework Equations


evaluating logarithms,
power of logarithms?

The Attempt at a Solution



SOLUTION ONE

=logxxn
=n(logxx)
=n(1)
=nSOLUTION TWO

=logxxn
=(logxx)n
=(1)n
=1

Please help me understand what I've done wrong. Thanks!
 
Last edited:
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the logx x^n means logx(x^n) not logx(x)^n so your solution two is wrong.

the logx(x^n) is the same as n * logx(x)
 
Tyrannosaurus_ said:
SOLUTION TWO

=logxxn
=(logxx)n
=(1)n
=1

Please help me understand what I've done wrong. Thanks!

Let ##y = \log_x x^n##. Then ##x^y = x^n \Rightarrow y = n##, so your solution one is correct.
##x^n## is in the argument of the log, the log is not to the nth power.
 
The other 2 posts are correct:

Your solution 1 uses logx(xn)
Your solution 2 uses (logx(x))n

An easy way to avoid this is to use brackets whenever applying a function:
 

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