# Value growing

• October 10th 2011, 01:52 PM
Greenid
Value growing
A value grows exponentially with a fixed percent per years. The value after 14 and 20 years have grown to 1819 and 6839. Determine the percentage it is growing per year.
Is there an equation for this or something? or do i have to guess and see if it end up at 1819 after 14 years? ..
• October 10th 2011, 02:29 PM
e^(i*pi)
Re: Value growing
Quote:

Originally Posted by Greenid
A value grows exponentially with a fixed percent per years. The value after 14 and 20 years have grown to 1819 and 6839. Determine the percentage it is growing per year.
Is there an equation for this or something? or do i have to guess and see if it end up at 1819 after 14 years? ..

The word exponentially means you're going have $e$ in there.

The general equation for exponential growth is $A(t) = A_0e^{kt}$ where:

• $A(t)$ = Value at time t
• $A_0$ = Initial value (t=0)
• $k$ = growth constant
• $t$ = time

In your case sub (14,1819) and (20,6839) for $(t,A(t))$ respectively into $A(t) = A_0e^{kt}$

You will then have a simultaneous equation with $A_0$ and $k$ as your unknowns. The question wants you to find k
• October 10th 2011, 03:32 PM
Greenid
Re: Value growing
Not sure if i understood right.
But, A(t) = 6839-1819 =5020
A0 = 1819
t = 6 ?
And then do ln 5020 = 1819 * k * ln e^6
k= ln 5020 / 1819 * ln e^6 ?
Kinda know this is wrong since my answer become so weird..
• October 10th 2011, 03:42 PM
skeeter
Re: Value growing
$6839 = A_0 e^{20k}$

$1819 = A_0 e^{14k}$

divide the first equation by the second ...

$\frac{6839}{1819} = e^{6k}$

solve for $k$
• October 10th 2011, 03:54 PM
Greenid
Re: Value growing
don't kill me, but no ln?
• October 10th 2011, 03:58 PM
skeeter
Re: Value growing
Quote:

Originally Posted by Greenid
don't kill me, but no ln?

what do you mean "no ln" ?
• October 10th 2011, 04:31 PM
e^(i*pi)
Re: Value growing
Quote:

Originally Posted by Greenid
don't kill me, but no ln?

I don't understand? Are you trying to solve this question without using the natural logarithm? If so that cannot be done AFAIK

Your next step is to take skeeter's equation from post 4 and take the natural logarithm of both sides noting that $\ln(e^{f(x)}) = f(x)$
• October 11th 2011, 05:04 AM
Greenid
Re: Value growing
I was just abit confused :)
ln e^6k = ln (6839/1819)
6k/6 = 1.3242/6
k= 0.2207 *100 = 22.07% ?
Correct now?
• October 11th 2011, 05:47 AM
e^(i*pi)
Re: Value growing
Quote:

Originally Posted by Greenid
I was just abit confused :)
ln e^6k = ln (6839/1819)
6k/6 = 1.3242/6
k= 0.2207 *100 = 22.07% ?
Correct now?

Yep. Looks good to me although you may want to put 22.07% per year
• October 11th 2011, 07:37 AM
Wilmer
Re: Value growing
Not sure what's going on here...but the "annual equivalent rate" seems to be
what's asked for; so it matters not HOW (weekly,daily, hourly, continuously...)
the interest is calculated; we have an opening and ending balance; so:
r = (6839/1819)^(1/6) - 1 = ~.24698 or ~24.7 %