Am I allowed to do this?

**ImaCowOK** · October 20th 2010, 02:22 PM

Put into Cartesian form with x>0

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

I solved for t:

$t=ln_e(x/5)$

Found the Cartesian:

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

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):

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

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

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

Divide both sides by $ln_e$

(yx)/5 = 3

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

Final answer 15/x

The correct answer is 15/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.

**Plato** · October 20th 2010, 02:31 PM

If you have more than one character in an exponent, set off the whole exponent in braces.
[tex]e^{-t}[/tex] gives $e^{-t}$ instead of $e^- t$ .

**skeeter** · October 20th 2010, 02:36 PM

$x = 5 e^t$

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

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

$3e^{-t} = 3 \cdot \frac{5}{x} = \frac{15}{x}$

**ImaCowOK** · October 20th 2010, 02:53 PM

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.

**Plato** · October 20th 2010, 03:08 PM

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.

**ImaCowOK** · October 20th 2010, 03:12 PM

Okai, thanks cupcake.

**Defunkt** · October 20th 2010, 03:26 PM

Originally Posted by ImaCowOK

Okai, thanks cupcake.

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

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

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

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

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

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