Exponential Growth (compounding) between two known values?

[tomadom]

I have two values and need to fill in the gaps. It's like the below.
I want to find what each value is as if it was continuously compounded so that by period 12 the value equals 33.

I've done a formula like this.

rate = (33/20)-1
period = 1/12 (for each day)

So for each day I calculate the value like this.

[The Day Before value] x (1+rate)^(1/12) but this doesn't seem to work.

How do I do this math problem? Also.... how does using exponential growth fit into this using e^rx7

1- 20
2 - ?
3 - ?
4 - ?
5 - ?
6 - ?
7 - ?
8 - ?
9 - ?
10 - ?
11 - ?
12 - 33

Thanks

 

[stapel]

I have two values and need to fill in the gaps. It's like the below.
I want to find what each value is as if it was continuously compounded so that by period 12 the value equals 33.

"Continuously compounded" means you'll be using the continuous-compounding formula from algebra (here):

. . . . .\(\displaystyle A\, =\, P\, e^{rt}\)

...where:

. . . . .A: ending amount
. . . . .P: starting amount
. . . . .r: compounding rate
. . . . .t: time

You say that the time periods are twelve, P = 20, and A = 33. So plug in, like you did back in algebra:

. . . . .\(\displaystyle 33\, =\, 20\, e^{12r}\)

Solve for the value of r. (here) Then use that value, along with the values of t for the various days, to fill in the table.

 

[HallsofIvy]

"Exponential growth" does not necessarily mean e to a power. Any thing of the form "\(\displaystyle a^t\)" for "a" greater than 1 gives "exponential growth". Of course, for any a, \(\displaystyle a^0= 1\) so I would start with \(\displaystyle Ba^{t- 1}\) so that when t= 1, this is \(\displaystyle Ba^0= B\) and the first condition just gives \(\displaystyle B= 20\). Now, set t= 12: \(\displaystyle 20a^{12-1}= 20a^{11}= 33\). You can solve that for a.