2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ?

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- Aug 28th 2012, 03:07 AMmathsgirl1234312e^2.1
2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ? - Aug 28th 2012, 03:13 AMProve ItRe: 2e^2.1
Hint: $\displaystyle \displaystyle \begin{align*} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \end{align*}$

Substitute $\displaystyle \displaystyle \begin{align*} x = 2.1 \end{align*}$ and evaluate to as much accuracy as you desire. - Aug 28th 2012, 03:59 AMemakarovRe: 2e^2.1
- Aug 28th 2012, 04:03 AMmathsgirl123431Re: 2e^2.1
Thanks guys, I appreciate the help

- Aug 28th 2012, 10:06 AMHallsofIvyRe: 2e^2.1
If it were me, I would use a calculator!

- Aug 28th 2012, 09:30 PMVlasevRe: 2e^2.1
Here's another perfectly fine way of doing this. You know how to take integer powers and suppose you have $\displaystyle e$ handy, to good precision. What can we do then? We have

$\displaystyle 2e^{2.1} = 2e^2 e^{1/10}$

Now you can use the Taylor series for that last exponential

$\displaystyle e^{1/h} = 1+\frac{1}{h}+\frac{1}{2h^2}+\frac{1}{6}h^3 + \cdots$

If you have $\displaystyle 2e^2$ to several decimal places in precision, then you can get away with using less terms in this series. It's not a big save but it's kind of neat. - Aug 29th 2012, 02:54 AMWilmerRe: 2e^2.1
- Aug 31st 2012, 05:55 AMnorthdownRe: 2e^2.1
I have finished up with the following and I want to find t;

2.6 = (e^-0.2t) + (e^-0.5t)

Any help and pointers would be appreciated. Thank you - Aug 31st 2012, 06:02 AMemakarovRe: 2e^2.1
Denote x = e^(-0.1t). Then your equation is 2.6 = x^2 + x^5. This will probably have to be solved numerically.