Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2018-01-10 22:10:16

Trigtrigger
Guest

### Exponential problem

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

Thank you

## #2 2018-01-11 01:04:35

bob bundy
Registered: 2010-06-20
Posts: 8,371

### Re: Exponential problem

hi Trigtrigger

Welcome to the forum.

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

Please see next post by Alg Num Theory.  I'll leave mine as a testament to the importance of 'reading the question' !!!

So 11.2 = a^(-0.85)

If you take logs (any will do but let's use base 10 as it's handy on a calculator) you get

log(11.2) = log(a^-0.85) = -0.85 * log(a)

So re-arrange this to make log(a) the subject, and inverse log it (ie. raise 10 to the power of) and you'll get a.

eg.  If log(a) = 2 then a = 10^2 = 100

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

## #3 2018-01-11 05:06:46

Alg Num Theory
Member
Registered: 2017-11-24
Posts: 300
Website

### Re: Exponential problem

bob bundy wrote:

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

So 11.2 = a^(-0.85)

Last edited by Alg Num Theory (2018-01-12 16:14:28)

Offline

## #4 2018-01-11 20:57:27

bob bundy
Registered: 2010-06-20
Posts: 8,371

### Re: Exponential problem

Thanks Alg Num Theory.  I seem to have lost an 'e'.  Thanks for the correction.  I have edited my post as a penance.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

## #5 2018-01-12 01:14:22

phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,600

### Re: Exponential problem

Alg Num Theory wrote:

...and if e is the constant that is Euler's number, then a's approximate value can be found to umpteen decimal places.

Also, y=ae^(-k*x) is different from the OP's y=ae^-k*x: the two forms don't give the same answer...as shown by MIF's calculator. Btw, the calculator treats 'e' as being Euler's number.

Last edited by phrontister (2018-01-13 20:48:09)

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline