Stuck doing parametric natural log graphs

[Witcher]

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.

 

[BvU]

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 $e^{ab} = e^{ba}$

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

 

[HallsofIvy]

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.

 

[Witcher]

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

 

[Mark44]

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.

 

[SammyS]

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 .

 

[Witcher]

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

Thanks.

 

[SammyS]

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.)

 

[archaic]

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