Proving the Property of Logarithms: Examples with Exponents

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The discussion revolves around the application of logarithmic properties to differentiate functions involving exponents. The original poster (OP) explores the transformation of y = x^{2/x} using logarithms, initially misapplying the property. Other participants clarify that the correct transformation is ln y = (2 ln x)/x, not (2/x)x, identifying the OP's mistake as a typo. The conversation concludes with the OP acknowledging the error. This highlights the importance of accuracy when applying logarithmic properties in calculus.
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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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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