2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ?

Printable View

- 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:

Substitute 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 handy, to good precision. What can we do then? We have

Now you can use the Taylor series for that last exponential

If you have 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.