- #1

- Thread starter Witcher
- Start date

- #1

- #2

BvU

Science Advisor

Homework Helper

2019 Award

- 13,424

- 3,197

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 got stuck

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

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

- #4

- 15

- 3

You can keep “e^t” and isolate t without using logrithms?

- #5

Mark44

Mentor

- 33,741

- 5,432

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.You can keep “e^t” and isolate t without using logrithms?

- #6

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,315

- 1,006

Hello, @Witcher . I see that you've been a member for a couple of months, but why not give you a welcome?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.

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:

- #7

- 15

- 3

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

Thanks.

Thanks.

- #8

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,315

- 1,006

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

Thanks.

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.)

- #9

- 562

- 164

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

- Last Post

- Replies
- 15

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 955

- Last Post

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 12

- Views
- 8K

- Last Post

- Replies
- 10

- Views
- 2K

- Last Post

- Replies
- 10

- Views
- 4K

- Last Post

- Replies
- 10

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 4K

- Last Post

- Replies
- 5

- Views
- 1K