# Stuck doing parametric natural log graphs

## Homework Statement:

I haven’t done logs in a few month and let alone with parametric graphs. I am having trouble with this problem. #35

## Relevant Equations:

X=e^t, y=e^3t
I got stuck when i eliminated the parameter.

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BvU
Homework Helper
2019 Award
I got stuck
in full sight of the harbour as they say in shipping language: ##y = e^{3\log x}## should remind you of something like ##e^{ab} = e^{ba}##

I use ##\log## for e based logarithms. Only engineers confuse e and 10, which is why they need ##\log## and ##\ln## .

HallsofIvy
Homework Helper
You have $x= e^t$ and $y= e^{3t}= (e^t)^3$ so $y= x^3$. I don't see any reason to use logarithms.

Chestermiller, berkeman, Witcher and 1 other person
You can keep “e^t” and isolate t without using logrithms?

Mark44
Mentor
You can keep “e^t” and isolate t without using logrithms?
Yes, because you don't need to isolate t. As has already been explained, ##e^{3t} = (e^t)^3##, so you can write y in terms of x, getting rid of the parameter t.

Witcher and berkeman
SammyS
Staff Emeritus
Homework Helper
Gold Member
Homework Statement:: I haven’t done logs in a few month and let alone with parametric graphs. I am having trouble with this problem. #35
Homework Equations:: X=e^t, y=e^3t

I got stuck when i eliminated the parameter.
Hello, @Witcher . I see that you've been a member for a couple of months, but why not give you a welcome?

You have been led to and/or given shorter ways to the answer, but your start was OK.

Recall that ##\ \ C\cdot \ln(x) = \ln(x^C) ##.

Apply that to ##\ \ 3(\ln(x)) ##, and proceed .

Last edited:
Witcher
I get it now but it wasn’t easy, my instinct was to Ln both sides when i seen the e

Thanks.

BvU
SammyS
Staff Emeritus
Homework Helper
Gold Member
I get it now but it wasn’t easy, my instinct was to Ln both sides when i seen the e.

Thanks.
As I mentioned, the path you started down was fine. It makes sense to work with the logarithm rules you may currently be studying and/or those rules you are most familiar with.

Carrying on from where you left off, (with ##\displaystyle y=e^{3(\ln(x))} ##):

You then have ##\displaystyle y=e^{\ln(x^3)} ##.

The final result follows immediately. (I hope.)

One can also use the fact that ##a^{bc}=(a^b)^c##.