Proving the Property of Logarithms: Examples with Exponents

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Homework Help Overview

The discussion revolves around properties of logarithms, specifically focusing on manipulating logarithmic expressions involving exponents. The original poster explores the application of logarithmic properties to a function of the form y = x^{2/x}.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply logarithmic properties to a new function and questions the correctness of their manipulation. Some participants question the accuracy of the original poster's expression and suggest it may contain a typo.

Discussion Status

The discussion includes attempts to clarify the logarithmic manipulation and identify potential errors. While some participants suggest that the original poster may have resolved their query, there is still an ongoing examination of the expressions involved.

Contextual Notes

There is a mention of a possible typo in the original poster's manipulation of the logarithmic expression, which may affect the understanding of the problem. The discussion reflects a learning environment where participants are checking assumptions and clarifying definitions.

Nano-Passion
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Edit: I answered my own question, I guess this thread serves no purpose so mods, you can delete this.

y = x^2
ln y = ln x^2
ln y = 2 ln x

Can we do the same thing with:

y = x^{2/x}
ln y = ln x^{2/x}
ln y = \frac{2}{x} x

Would that be correct? I just want to make sure because I used this technique for differentiation.
 
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Doesn't: ln y = \frac{2}{x} x just make it 2 :p

But from what I remember about logs, yes you can do that..
 
The OP might have already figured this out, but ln x^{2/x} = \frac{2 ln(x)}{x}, not \frac{2}{x}x. Typo maybe?
 
gb7nash said:
The OP might have already figured this out, but ln x^{2/x} = \frac{2 ln(x)}{x}, not \frac{2}{x}x. Typo maybe?

Yep, it was a typo.
 

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