Question 637300
You're not very confused because everything you figured out is correct. The only thing you're missing is how to find the "k".<br>
To find the "k", you use the equation, {{{A = A[0]*e^(kt)}}}, and the information you were given about the sales in 2002. In the equation...
A = the sales t years after the starting point, 2000
{{{A[0]}}} is the sales at the starting point, 2000. IOW, {{{A[0]}}} is the amount at t = 0.
e = Euler's constant
k = some constant
t = a number of years after the starting point
With the information you've been given, our
{{{A[0] = 10000}}}
For the year 2002
A = 82700
t = 2 (Since 2002 is two years after 2000)
This makes our working equation:
{{{82700 = 10000*e^(k*2)}}}
We can solve this for k. Dividing both sides by 10000:
{{{8.27 = e^(k*2)}}}
Find the log of each side, using ln because of the "e":
{{{ln(8.27) = ln(e^(k*2))}}}
Using a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}} we can move the exponent out in front:
{{{ln(8.27) = (k*2)*ln(e)}}}
Since ln(e) = 1 this simplifies to:
{{{ln(8.27) = (k*2)}}}
Dividing by 2 we get:
{{{ln(8.27)/2 = k}}}
This is an exact expression for k. We probably want to use a decimal so from our calculator we should get:
1.05631725 = k<br>
Now that we have the k, we have a full equation:
{{{A = 10000*e^(1.05631725t)}}}
which can then use this equation to answer other questions.<br>
To find when the sales hit 500000:
{{{500000 = 10000*e^(1.05631725t)}}}
Now solve this for t just like we solved the earlier equation for k. You will get a decimal answer. Let's pretend you get t = 5.2117. Since 5.2117 is more than 5, the sales would reach 500000 more than 5 years after the start. IOW, the sales would reach 500000 sometime during the sixth year: 2006. (Note: 5.2117 is a pretend answer. You will not actually get 5.2117 for an answer.)