# Exponential problem

• Jun 21st 2010, 03:36 AM
bbeweel
Exponential problem
A exponential curve relating two quantities 'q' and 't' has the form q=ae^kt where a and k are constants.
The curve passes through two points (0.09036,2) and (3.307,-4)
Determine the values of 'a' and 'k' and find the value of 't' when q= 1.5

I have no idea where to even start this one guys could someone help me.
• Jun 21st 2010, 03:42 AM
Prove It
$\displaystyle q = a\,e^{k\,t}$, and you know that $\displaystyle (t, q) = (0.09036, 2)$ and $\displaystyle (t, q) = (3.307, -4)$ lie on the curve.

What you have been given in nonsense, as the exponential function never crosses the horizontal axis. So you can't have the values of $\displaystyle q$ being different sign...

Just to be sure:

Substituting these points gives two equations in two unknowns:

$\displaystyle 2 = a\,e^{0.09036k}$

$\displaystyle -4 = a\,e^{3.307k}$.

Dividing Equation 2 by Equation 1 gives:

$\displaystyle \frac{-4}{\phantom{-}2} = \frac{a\,e^{3.307k}}{a\,e^{0.09036k}}$

$\displaystyle -2 = e^{3.21664k}$.

This is clearly impossible since the exponential function is always positive.
• Jun 22nd 2010, 02:24 AM
bbeweel
Hi thanks for replying,you say the problem is nonsense but i have worked it to this below,have i not understood the problem ?

q= ae^kt

3.307=ae^ -4k

0.09036=ae^2k

36.598=e^ -6k

-6k=ln36.598 = 3.599

k= 3.599/-6 = -0.599

3.307=ae^2.396

a=3.307/e^2.396 = 0.3012

when q = 1.5

1.5 = 0.3012e^-0.599t

1.5/0.3012 = e^-0.599t

4.98 = e^-0.599t

-0.599t = ln4.98 = 1.6054

t= 1.6054/-0.599 = -2.68
• Jun 22nd 2010, 06:02 AM
Quote:

Originally Posted by bbeweel
Hi thanks for replying,you say the problem is nonsense but i have worked it to this below,have i not understood the problem ?

q= ae^kt

3.307=ae^ -4k

0.09036=ae^2k

36.598=e^ -6k

-6k=ln36.598 = 3.599

k= 3.599/-6 = -0.599

3.307=ae^2.396

a=3.307/e^2.396 = 0.3012

when q = 1.5

1.5 = 0.3012e^-0.599t

1.5/0.3012 = e^-0.599t

4.98 = e^-0.599t

-0.599t = ln4.98 = 1.6054

t= 1.6054/-0.599 = -2.68

you have messed up the 'q' and 't' while substituting and how can time be negative?
• Jun 22nd 2010, 06:52 AM
bbeweel
Yes i overlooked time being negative :-/
I have swapped my 'q' and 't' around and i now have the answers k = 0.600 , a = 0.996 and t = 0.6823 ??? could you please let me know if i am even close here ?
• Jun 22nd 2010, 07:08 AM
skeeter
Quote:

Originally Posted by bbeweel
Yes i overlooked time being negative :-/
I have swapped my 'q' and 't' around and i now have the answers k = 0.600 , a = 0.996 and t = 0.6823 ??? could you please let me know if i am even close here ?

why are you swapping dependent and independent variables? did you make a mistake interpreting the original problem statement?

it would help if you stated the original problem in its entirety.
• Jun 22nd 2010, 07:11 AM
bbeweel
Quote:

Originally Posted by skeeter
why are you swapping dependent and independent variables? did you make a mistake interpreting the original problem statement?

it would help if you stated the original problem in its entirety.

A exponential curve relating two quantities 'q' and 't' has the form q=ae^kt where a and k are constants.
The curve passes through two points (0.09036,2) and (3.307,-4)
Determine the values of 'a' and 'k' and find the value of 't' when q= 1.5

I have worked it to this so far below:

q= ae^kt

3.307=ae^ -4k

0.09036=ae^2k

36.598=e^ -6k

-6k=ln36.598 = 3.599

k= 3.599/-6 = -0.599

3.307=ae^2.396

a=3.307/e^2.396 = 0.3012

when q = 1.5

1.5 = 0.3012e^-0.599t

1.5/0.3012 = e^-0.599t

4.98 = e^-0.599t

-0.599t = ln4.98 = 1.6054

t= 1.6054/-0.599 = -2.68
• Jun 22nd 2010, 08:04 AM
skeeter
Quote:

Originally Posted by bbeweel
A exponential curve relating two quantities 'q' and 't' has the form q=ae^kt where a and k are constants.
The curve passes through two points (0.09036,2) and (3.307,-4)
Determine the values of 'a' and 'k' and find the value of 't' when q= 1.5

the ordered pair is $\displaystyle (t,q)$

as stated earlier, this problem is bogus. the only way to get a negative value for $\displaystyle q$ is for $\displaystyle a$ to be negative ... but then, all values of $\displaystyle q$ would have to be negative.

if the equation has the form $\displaystyle q = ae^{kt} + c$ , then a solution may exist with the given coordinates.

btw ... time can be negative, it's just a time prior to when t = 0.

if the coordinates were swapped, then your solution is close.

I get $\displaystyle k = -0.5999$ , $\displaystyle a = 0.3000$ , and $\displaystyle q(1.5) = 0.1220$