Is the derivative of log base 3 of e^2x correct?

[Jan Hill]

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

 

[vela]

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.

 

[Jan Hill]

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 ?

 

[vela]

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.

 

[Mark44]

This can be done in a much simpler way, using the properties of logarithms.

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