SOLUTION: Find an exponential Formula for f(-8) = 200 and f(30) = 580. Could you please explain to me what process I use to find the answer to this kind of problem? I wish to find the answe

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find an exponential Formula for f(-8) = 200 and f(30) = 580. Could you please explain to me what process I use to find the answer to this kind of problem? I wish to find the answe      Log On


   

Question 84434: Find an exponential Formula for f(-8) = 200 and f(30) = 580.
Could you please explain to me what process I use to find the answer to this kind of problem? I wish to find the answer to the problem after understanding the process if I can. I don't wish for the answer, just an explanaton on how to find the answer if that is okay. I was able to figure out that I need to do something to show that there are 38 units between f(-8) and f(30) and I even found the answer using the calculator and lots of trial and error, but I simply can't figure out how to solve it properly. Please help.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given f(-8) = 200 and f(30) = 580
.
Find an exponential formula that fits these given data.
.
The formula you will need to find is of the form +y+=+A%2Ae%5E%28b%2Ax%29
.
So what you need to do is to solve for A and b using the given data and appropriate
math processes.
.
Start with the given that y = 200 when x = -8. Plug these values into the formula to get:
.
200+=+A%2Ae%5E%28b%2A%28-8%29%29
.
Take the natural logarithm of both sides:
.
ln%28200%29+=+ln%28A%2Ae%5E%28-8%2Ab%29%29+=+ln%28A%29+-8%2Ab%2Aln%28e%29
.
But the natural log of e is 1. Substitute that and you get:
.
ln%28200%29+=+ln%28A%29-8%2Ab+
.
Solve this for ln(A) by adding 8*b to both sides to get:
.
ln%28A%29+=+ln%28200%29%2B+8%2Ab
.
Next follow the identical process for the second data point in which y = 580 and x = 30.
Using this data in the formula and solving for ln(A) results in:
.
ln%28A%29+=+ln%28580%29+-+30%2Ab
.
Now you can set the right sides of the two equations you have for ln(A) equal and solve
for b. You should be solving:
.
ln%28200%29+%2B+8%2Ab+=+ln%28580%29+-+30%2Ab
.
When you solve this you should find that b is somewhere around 0.028018703 or whatever
you choose to round it to.
.
Now that you know b you can return to either of the two equations for the data points and
solve for A. For example, return to:
.
y+=+A%2A%28e%5E%28b%2Ax%29%29
.
Substitute 200 for y, -8 for x, and 0.028018703 for b to get:
.
200+=+A%2A%28e%5E%280.028018703%2A%28-8%29%29%29=+A%2Ae%5E%28-0.224149628%29
.
Solve for A by dividing both sides by e%5E%28-0.224149628%29
.
A+=+200%2F%28e%5E%28-0.224149628%29%29
.
But a quantity in the denominator that has a negative exponent can be brought into the
numerator with the same exponent only positive. This means that A becomes equal to:
.
A+=+200%2Ae%5E%280.224149628%29
.
and by a calculator e%5E%280.224149628%29+=+1.25125823. Substituting this into the equation
for A results in:
.
A+=+200%2A1.25125823+=+250.2516459
.
Now put this value of A and also the known value of b into the equation and you should
end up with something around:
.
y+=+%28250.2516459%29%2Ae%5E%280.028018703%2Ax%29
.
as the equation you are looking for.
.
Hope you can track your way through this. You can check by letting x = 30 and solving
to see if y = 580. You should find that y computes to be 579.9999895 or so. Close
enough to assume this works. Cheers and good luck!