# Am I allowed to do this?

• Oct 20th 2010, 02:22 PM
ImaCowOK
Am I allowed to do this?
Put into Cartesian form with x>0

$\displaystyle x=5e^t$ $\displaystyle y=3e^{-t}$

I solved for t:

$\displaystyle t=ln_e(x/5)$

Found the Cartesian:

$\displaystyle y=3e^{-ln_e(x/5)}$

This is how I got the final answer of 15/x and I'd like to know if, mathematically, I'm allowed to do it this way(I might leave out some steps, but I think you will understand. Let me know if you want me to add those steps):

$\displaystyle y=3/e^{ln(x/5)}$

Multiply both sides by $\displaystyle e^{ln(x/5)}$

Change to the form:$\displaystyle yln_e(x/5) = ln_e3$

Divide both sides by $\displaystyle ln_e$

(yx)/5 = 3

Multiply both sides by 5, then divide both sides by x.

Sorry about not using MATH tags all the way through. I'm under time constraints and am not very familiar/savvy with MATH tags.
• Oct 20th 2010, 02:31 PM
Plato
If you have more than one character in an exponent, set off the whole exponent in braces.
$$e^{-t}$$ gives $\displaystyle e^{-t}$ instead of $\displaystyle e^- t$.
• Oct 20th 2010, 02:36 PM
skeeter
$\displaystyle x = 5 e^t$

$\displaystyle \frac{x}{5} = e^t$

$\displaystyle \frac{5}{x} = e^{-t}$

$\displaystyle 3e^{-t} = 3 \cdot \frac{5}{x} = \frac{15}{x}$
• Oct 20th 2010, 02:53 PM
ImaCowOK
I don't understand what you did there. Just posting that without an explanation doesn't help. Was the way I solved for the final answer wrong? That's what I'm asking. Are you showing a simpler and quicker way to do it? Those are welcome too, but I need to understand what you did to get there.
• Oct 20th 2010, 03:08 PM
Plato
Quote:

Originally Posted by ImaCowOK
Was the way I solved for the final answer wrong?

What you did is correct.
It is a matter of choice the way to do it.
• Oct 20th 2010, 03:12 PM
ImaCowOK
Okai, thanks cupcake.
• Oct 20th 2010, 03:26 PM
Defunkt
Quote:

Originally Posted by ImaCowOK
Okai, thanks cupcake.

Your solution is actually partially correct - you don't actually "divide by $\displaystyle ln_e$" because it is not a number.

Also, $\displaystyle ln$ denotes the logarithm to the base of $\displaystyle e$, so writing $\displaystyle ln_e$ is redundant.

Generally, if you have $\displaystyle lnx = lny$ then you can raise e to both sides of the equation to get $\displaystyle e^{lnx} = e^{lny}$ and since $\displaystyle e^{lnx} = ln(e^x) = x$, you get that $\displaystyle x=y$.

Note that you could've actually finished in the third line using logarithm \ exponent rules:

$\displaystyle y=3e^{-ln(x/5)} = 3e^{ln(\frac{5}{x})} = 3 \cdot \frac{5}{x} = \frac{15}{x}$