# e^-x = 1/e^x proof?

• May 28th 2010, 03:15 AM
gerardhoyle
e^-x = 1/e^x proof?
Hi I've been asked to prove $\displaystyle e^{-x} = \frac{1}{e^x}$

using $\displaystyle e^{x_1} \cdot e^{x_2} \ = \ e^{x_1 + x_2}$

but I'm not sure how to go about it.

If I start;

$\displaystyle y_1 = e^{x_1}$
$\displaystyle y_2 = e^{x_2}$

$\displaystyle \ln y_1 = x_1$
$\displaystyle \ln y_2 = x_2$

$\displaystyle x_1 + x_2 = \ln y_1 + \ln y_2$
$\displaystyle x_1 + x_2 = \ln (y_1 y_2)$

I don't see how to go anywhere else but to prove that

$\displaystyle e^{x_1} \cdot e^{x_2} = e^{x_1 + x_2}$

(The latex is really messed up on this site & I can't put minuses & subscripts into the exponents, I've made it as reasonably clear as I can).

Moderator edit: No, the latex is not "really messed up on this site". You just don't know how to format the code. Reading the Latex Tutorial always helps: http://www.mathhelpforum.com/math-he...-tutorial.html.
• May 28th 2010, 03:18 AM
Prove It
Quote:

Originally Posted by gerardhoyle
Hi I've been asked to prove $\displaystyle e^-x = \frac{1}{e^x}$

using $\displaystyle e^x_1 \cdot e^x_2 \ = \ e^x_1 + ^x_2$

but I'm not sure how to go about it.

If I start;

$\displaystyle y_1 = e^x1$
$\displaystyle y_2 = e^x2$

$\displaystyle lny_1 = x_1$
$\displaystyle lny_2 = x_2$

$\displaystyle x_1 + x_2 = lny_1 + lny_2$
$\displaystyle x_1 + x_2 = ln(y_1y_2)$

I don't see how to go anywhere else but to prove that

$\displaystyle e^x_1 \cdot e^x_2 = e^x_1 + ^x_2$

(The latex is really messed up on this site & I can't put minuses & subscripts into the exponents, I've made it as reasonably clear as I can).

It's really quite simple...

Use the index law $\displaystyle a^{-n} = \frac{1}{a^n}$.

So $\displaystyle e^{-x} = \frac{1}{e^{x}}$.
• May 28th 2010, 03:24 AM
gerardhoyle
Yeah I really think it's just that, however the book is telling me to use

$\displaystyle e^x \cdot e^y = e^x + ^y$

(I just use y instead of x_2 to ease up on the latex).

& I can't see how it comes about. These proof's use technique's like 'functions with the same derivative differ by a constant' & logarithmic differentiation to prove some of these laws, I don't see how any of it fits into this.
• May 28th 2010, 03:27 AM
HallsofIvy
Note that to have more than one symbol in an exponent (or numerator of a fraction or denominator of a fraction or in a sqrt symbol...) you need to enclose them in { }.

I think you are saying that you need to show that $\displaystyle e^{-x}= \frac{1}{e^x}$, using the fact that $\displaystyle a^{x_1}a^{x_2}= a^{x_1+ x_2}$

Don't use logarithms, just look at $\displaystyle e^xe^{-x}= e^{x+ (-x)}= e^0= 1$.
• May 28th 2010, 03:37 AM
gerardhoyle
Quote:

Originally Posted by HallsofIvy
Note that to have more than one symbol in an exponent (or numerator of a fraction or denominator of a fraction or in a sqrt symbol...) you need to enclose them in { }.

Oh man, thank you! You have no idea how frustrating that has been (Rofl)

The conclusion I draw from that is;

$\displaystyle 1 = e^0 = e^{x + (-x)} = e^x \cdot e^{-x} = e^x \cdot \frac{1}{e^x} = \frac{e^x}{e^x} = 1$

but I don't know how satisfying that is.
• May 28th 2010, 04:24 AM
gerardhoyle
And if that proves that $\displaystyle e^x = \frac{1}{e^x}$ then I could use this fact to prove:

$\displaystyle \frac{e^x}{e^y} = e^{x - y}$

by

$\displaystyle e^{x - y} = e^x \cdot e^{-y} = e^x \cdot \frac{1}{e^y} = \frac{e^x}{e^y}$

Then smushing it all together to prove:

$\displaystyle (e^{x_1})^{x_2} = e^{x_1x_2}$

by;

$\displaystyle y_1 = e^{x_1}$

$\displaystyle y_2 = y_1^{x_2}$

$\displaystyle lny_1 = x_1$

$\displaystyle lny_2 = x_2lny_1 = x_1x_2$

$\displaystyle lny_2 = x_1x_2$

$\displaystyle y_2 = e^{x_1x_2}$

(Nerd)

(Thanks, if there's no error here then thanks for pointing me in the right direction, darn book!).
• May 28th 2010, 06:37 PM
AllanCuz
Quote:

Originally Posted by gerardhoyle
And if that proves that $\displaystyle e^x = \frac{1}{e^x}$ then I could use this fact to prove:

$\displaystyle \frac{e^x}{e^y} = e^{x - y}$

by

$\displaystyle e^{x - y} = e^x \cdot e^{-y} = e^x \cdot \frac{1}{e^y} = \frac{e^x}{e^y}$

Then smushing it all together to prove:

$\displaystyle (e^{x_1})^{x_2} = e^{x_1x_2}$

by;

$\displaystyle y_1 = e^{x_1}$

$\displaystyle y_2 = y_1^{x_2}$

$\displaystyle lny_1 = x_1$

$\displaystyle lny_2 = x_2lny_1 = x_1x_2$

$\displaystyle lny_2 = x_1x_2$

$\displaystyle y_2 = e^{x_1x_2}$

(Nerd)

(Thanks, if there's no error here then thanks for pointing me in the right direction, darn book!).

I don't know why Hallsofivy told you to stay away from logs but...

$\displaystyle \frac{e^x}{e^y} = e^{x - y}$

$\displaystyle ln \frac{e^x}{e^y} = ln e^{x - y}$

$\displaystyle lne^x - lne^y = (x-y)$

$\displaystyle x - y = x - y$

Seems simpler to me?

$\displaystyle (e^{x_1})^{x_2} = e^{x_1x_2}$

$\displaystyle ln (e^{x_1})^{x_2} = ln e^{x_1x_2}$

$\displaystyle x_2 ln e^{x_1} = x_1 x_2$

$\displaystyle x_2 x_1 = x_2 x_1$

Also to prove your original question with logs

$\displaystyle e^{-x} = \frac{1}{e^x}$

$\displaystyle lne^{-x} = ln \frac{1}{e^x}$

$\displaystyle -x = ln1 - lne^x$

$\displaystyle -x = -x$
• May 29th 2010, 12:54 AM
CaptainBlack
Quote:

Originally Posted by AllanCuz
I don't know why Hallsofivy told you to stay away from logs but...

He told the OP to stay away from logs because the solution takes three lines just using the exponent law given:

As $\displaystyle e^xe^y=e^{x+y}$ we have:

$\displaystyle 1=e^0=e^{x-x}=e^x e^{-x}$

Hence $\displaystyle e^{-x}=1/e^x$

CB
• May 29th 2010, 12:55 AM
CaptainBlack
Quote:

Originally Posted by AllanCuz
I don't know why Hallsofivy told you to stay away from logs but...

$\displaystyle \frac{e^x}{e^y} = e^{x - y}$

This assumes more than we are asked to prove so is not permitted, so for your argument to be valid you need to demonstrate that every step is reversible.

CB
• May 29th 2010, 01:40 AM
HallsofIvy
Quote:

Originally Posted by AllanCuz
I don't know why Hallsofivy told you to stay away from logs but...

I said that because the original post had said
"I've been asked to prove $\displaystyle e^{-x}= \frac{1}{e^x}$`

using $\displaystyle e^{x+ y}= e^xe^y$."
• May 29th 2010, 07:51 AM
AllanCuz
Quote:

Originally Posted by CaptainBlack
This assumes more than we are asked to prove so is not permitted, so for your argument to be valid you need to demonstrate that every step is reversible.

CB

What? That was the question: to prove that statement. If we prove the left side equals right side by reverse engineering, that's perfectly acceptable.
• May 29th 2010, 09:58 AM
gerardhoyle
Quote:

Originally Posted by gerardhoyle
Hi I've been asked to prove $\displaystyle e^{-x} = \frac{1}{e^x}$

using $\displaystyle e^{x_1} \cdot e^{x_2} \ = \ e^{x_1 + x_2}$

but I'm not sure how to go about it.

While I liked the way you did it AllanCuz I was asking for instructions on how to derive the identity explicitly Going off $\displaystyle e^{x_1} \cdot e^{x_2} \ = \ e^{x_1 + x_2}$ as I had already proved that.

Also, sorry about the latex comment, the use of latex is the exact same to that on other sites I've used in every way except putting more than one term in the exponent, I incorrectly assumed it was a latex problem as I've been on one site where there was an issue with latex.
• May 29th 2010, 11:18 AM
CaptainBlack
Quote:

Originally Posted by AllanCuz
What? That was the question: to prove that statement. If we prove the left side equals right side by reverse engineering, that's perfectly acceptable.

Only if every step is reversible (which it is) but you still have to claim it if not prove it (or better actualy reverse the argument)

CB