# How do I manually calculate e^x?

• Apr 23rd 2008, 05:49 AM
Hanga
How do I manually calculate e^x?
Okay so I've been calculating using the simple e^x derivations. I found that on my Texas TI-89 Titanium I can use the nDeriv( command to EXTREMLY easy and fast calculate pretty much any derivation as long as I have something to go at.

So how would you guys/girls calculate this;

I(7) if
I(x) = 15'000*e^(0.24x)

Is is possible to do this without "cheating" using my calc?
• Apr 23rd 2008, 05:57 AM
colby2152
Quote:

Originally Posted by Hanga

I(7) if
I(x) = 15'000*e^(0.24x)

Is is possible to do this without "cheating" using my calc?

Do you know to derive an exponential function using the chain rule? If not, then some of the lessons found here may help you.
• Apr 23rd 2008, 06:11 AM
Hanga
Quote:

Originally Posted by colby2152
Do you know to derive an exponential function using the chain rule? If not, then some of the lessons found here may help you.

I'm from Sweden so I don't know your english technical terms for math :)

Anyways I pretty much do know how to derivate it;

f(x) 15'000e^(0,24x)
f´(x) = 3600e^(0,24x)
f´(7) = 3600e^(0,24*7)

and so on :)
Still.. I can't calculate this without using my calc. How would you do it?
• Apr 23rd 2008, 06:16 AM
Isomorphism
Quote:

Originally Posted by Hanga
I'm from Sweden so I don't know your english technical terms for math :)

Anyways I pretty much do know how to derivate it;

f(x) 15'000e^(0,24x)
f´(x) = 3600e^(0,24x)
f´(7) = 3600e^(0,24*7)

and so on :)
Still.. I can't calculate this without using my calc. How would you do it?

You have to use the calculator... Or at least use what the calculator uses ;)
I mean use the power series definition to get it up to required accuracy.You know $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$, right?
• Apr 23rd 2008, 06:24 AM
Hanga
Quote:

Originally Posted by Isomorphism
You have to use the calculator... Or at least use what the calculator uses ;)
I mean use the power series definition to get it up to required accuracy.You know $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$, right?

Hehe im not in that grade of math just yet :)
Still, using my calculator to do my original math problem is not really wrong then?
• Apr 23rd 2008, 07:44 AM
CaptainBlack
Quote:

Originally Posted by Hanga
Hehe im not in that grade of math just yet :)
Still, using my calculator to do my original math problem is not really wrong then?

What definition of the exponential function or "e" do you have, and what do you know about it.?

RonL
• Apr 23rd 2008, 09:05 AM
Hanga
Quote:

Originally Posted by CaptainBlack
What definition of the exponential function or "e" do you have, and what do you know about it.?

RonL

Okay so using my original quesion
what is f´(7) if

f(x)=15'000e^(0,24x)

The answer will be 19316 which is close to 19300.
What I do know about e is that it's used to provide a simple derivation to
y = a^x
• Apr 23rd 2008, 09:49 AM
colby2152
Quote:

Originally Posted by Hanga
What I do know about e is that it's used to provide a simple derivation to
y = a^x

That's not true... I wrote some notes on exponentials that should be very helpful.

The power series that Iso provided is about the only way you could calculate such values without the use of a calculator, unless you can do such arithmetic in your head which is possible if your estimate the value of e to be about 2.7.
• Apr 23rd 2008, 02:07 PM
CaptainBlack
Quote:

Originally Posted by Hanga
Okay so using my original quesion
what is f´(7) if

f(x)=15'000e^(0,24x)

The answer will be 19316 which is close to 19300.
What I do know about e is that it's used to provide a simple derivation to
y = a^x

Then there is something wrong with the course you are pursuing, you are using things that as far as you are concerned can be little different from magic.

RonL
• Apr 23rd 2008, 03:31 PM
Mathstud28
Quote:

Originally Posted by Hanga
Okay so using my original quesion
what is f´(7) if

f(x)=15'000e^(0,24x)

The answer will be 19316 which is close to 19300.
What I do know about e is that it's used to provide a simple derivation to
y = a^x

If you are saying that you can use e to find ${D[a^{x}]}{dx+ln(a)a^{x}$ I suppose you say this extremely loosely considering their is a ln(a) in the derivative which is the inverse function of x??

and you could also graph it to approximate it if you dont want to use a power series
• Apr 23rd 2008, 04:21 PM
icemanfan
What he might be referring to in the "simple derivation" for y = a^x could be that the derivative of $a^x$ is $(\ln{a})a^x$ and you can't find the natural logarithm without e. Or I could be wrong.
• Apr 23rd 2008, 04:28 PM
Mathstud28
Quote:

Originally Posted by icemanfan
What he might be referring to in the "simple derivation" for y = a^x could be that the derivative of $a^x$ is $(\ln{a})a^x$ and you can't find the natural logarithm without e. Or I could be wrong.

Yeah that is what I said in the post directly above yours