1. 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.

2. $\displaystyle q = a\,e^{k\,t}$, 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.

3. 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

4. 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?

5. 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 ?

6. 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.

7. 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

8. 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 $\displaystyle q(1.5) = 0.1220$