# Finding constants in exponential functions

1. Jan 20, 2015

### LivvyS

1. The problem statement, all variables and given/known data
In 2000 the population of a country was estimated to be 8.23 million. In 2010 the population was 9.77 million.

Assume that the number of people P(t) in millions at time t (in years since 2000) is modelled by the exponential growth function.

P(t) = Aekt

Find P(t) giving the two constants in it to 2 significant figures.

2. Relevant equations
P(t) = Aekt

3. The attempt at a solution

P(0) = 8.23x106 so Ae0k= 8.23x106

P(10) = 9.77x106 so Ae10k= 9.77x106

Divide to eliminate A:
Ae0k= 8.23x106 / Ae10k= 9.77x106 = e0k-10k= 8.23x106 / 9.77x106 = e-10 k = 8.23x106 / 9.77x106

(I am not certain that this step is right)

-10k = ln 8.23x106 / 9.77x106

k = 0.01917945693

To find A:
Ae10k= 9.77x106
Ae10*0.01917945693= 9.77x106
A = 9.77x106 / e10*0.01917945693 = 8064904.714

These values for k and A seem to produce 9.77x106 when 10 (years) is plugged into the function, however I cant seem to produce 8.23x106 at 0 years since 2000 and I cant seem to see why, any help would be much appreciated!

2. Jan 20, 2015

### Quantum Defect

The trick is to remember what the value of the exponential is when t = 0. What is exp(0)? If you know what the population is at t=0, how is this related to A?

3. Jan 20, 2015

### LivvyS

So if the exponent of e is 0 then P(0) = A*1 so A = 8.23x106
But if this were true than the answer for a I calculated cannot be right, if so could you give me a clue as to whats wrong with it?

4. Jan 20, 2015

### BvU

Method is right, something goes wrong from -10k = ln 8.23x106 / 9.77x106 ⇒k = 0.01917945693

(you typed ... instead of ...)

5. Jan 20, 2015

### Ray Vickson

It is much, much easier just to work in units of millions, so that your two equations are
$$A = 8.23 \; \text{and}\; A e^{10 k} = 9.77$$
By choosing those new units you eliminate a lot of the clutter and make it much easier to see what is happening (and to find errors). You can always convert to other units later, after you have solved the problem.

6. Jan 21, 2015