Logarithm Derivative: Solving y = log base 3 e^2x with Homework Equations

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Homework Help Overview

The discussion revolves around finding the derivative of the function y = log base 3 of e raised to the power of 2x. Participants are exploring the application of logarithmic differentiation and the properties of logarithms in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to differentiate the function using logarithmic rules and the chain rule. There are questions about the correctness of the derivative obtained, with some participants suggesting simplifications and others expressing confusion about the steps involved.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants providing guidance on the differentiation process. Multiple interpretations of the derivative are being discussed, and while some corrections are suggested, there is no explicit consensus on the final form of the derivative.

Contextual Notes

Some participants mention difficulties in presenting mathematical expressions clearly, which may affect the clarity of the discussion. There is also a reference to using properties of logarithms to simplify the differentiation process.

Jan Hill
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Homework Statement



y = log base 3 e^2x

Homework Equations





The Attempt at a Solution



I got y' = 2e^2x divided by e^2xln3

is this right?
Sorry for the pathetic way of presenting this. I haven't been able to use the lancet program for proper ways to write mathematical stuff
 
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Jan Hill said:

Homework Statement



y = log base 3 e^2x

Homework Equations





The Attempt at a Solution



I got y' = 2e^2x divided by e^2xln3

is this right?
Sorry for the pathetic way of presenting this. I haven't been able to use the lancet program for proper ways to write mathematical stuff
So are you saying you got

y' = \frac{2e^{2x}}{e^{2x}\log 3} = \frac{2}{\log 3}

If so, that's not correct. Show us your work so we can see where you're going astray.
 
the derivative of e^2x is itself, e^2x and then this is multiplied by 2.

so that part is 2e^2x

and the derivative of any non ln logarithm is 1/ln of the base which gives me 1/ln3

maybe the answer should be 2e^2x/ln3 ?
 
I'm sorry! You had it right the first time. (For some reason, I kept thinking there should be an x floating around.) The only thing is you could have simplified your answer to get rid of the exponentials.
 
This can be done in a much simpler way, using the properties of logarithms.

y = log3 e2x = 2x * log3 e
==> y' = 2 * log3 e
 

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