exponential groth function

Hey there, Imani. I don't have much experience with exponential growth functions, but let me see if I can help at all.

The equation that I learned for an exponential growth function is :

y= a (1+r)^x

Where 'a' is the original amount, 'r' is the rate of growth (often a percent in decimal form), x is the number of time intervals, and y is the final answer. When the interest is compounded continuously the equation becomes:

y= a e^rt

y, a, r, and t all stay the same; and 'e' stands for Euler's number which is approximately 2.718281828 and can be used in equations like this to represent continuous growth or decay. There should be an 'e' button on your calculator that you should use to be more exact in you calculations than just "2.71828...." and it's easier than memorizing it.

Now you can plug your numbers into the equation and solve the problems.

a.) What is the exponential growth function in terms of P and 0.03?

'P' is the amount of money invested, so 'P' would be 'a' in our equation. and 0.03 would be the growth rate, 'r'.

y= p e^0.03t

(Please, someone correct me if I didn't read that problem right.)

b.) if $6000 is invested what is the balance after 4 years?

$6000 is the amount invested so it replaces 'a', 4 years is how long it was invested so it replaces 't'.

y= $6000 e^0.03(4)

I'll let you finish that equation.

c.) when will an investment of $6000 double itself?

For this, $12,000 ($6000x2) is the final answer so it goes is 'y's place. and the unknown in this problem is how many years, or 't'.

$12000 = $6000 e^0.03(t)

You can finish that one as well.

I hope that helped, let me know if I forgot something.

^_^

suppose that p is invested in a savings account in which interest k, is compounded continuously at 3% per year. What is the exponential growth function in terms of P and 0.03?
b) if $6000 is invested what is the balance after 4 years?
c) when will an investment of $6000 double itself?
------------------------------
I am going to re-word your question a little.
Suppose that P is invested in a savings account at 3% p.a. interest, compounding annually. (the interest is added at the end of each year)
I am going to call the future value S (in my experience S is the standard letter used for future value.)
I am going to let the number of years invested be n

a) S = P * 1.03 ⁿ

b) S = 6000 * 1.03 ⁴
you can finish this one.

c) 2P = P * 1.03 ⁿ
2 = 1.03 ⁿ
log 2 = log 1.03 ⁿ
log 2 = n*log1.03
n = log2 / log1.03
n = 23.449 years
Since the interest is added at the end of the year it will be doubled in 24 years (It will be a little more than double at this time)