Question

$1200 deposit into savings, earning 3% compounded continuously. Using A=Pe^rt determine balance after 30 years. also, how long does it take intital deposit to grow to $2000?

Answer 1

\[A = Pe^{rt}\]\[t = 30\]\[P=1200\]\[r=0.03\]
\[A = 1200(2.71828)^{0.03*30} =\]

For the second part,
\[A = Pe^{rt}\]\[\frac{A}{P} = e^{rt}\]\[\frac{\ln\frac{A}{P}}{r}=t\]

Answer 2

In the second part, be sure you calculate \[\ln \frac{A}{P}\]not \[\frac{\ln A}{\ln P}\]as I did while checking my work :-)

Answer 3

so A= 2951
and t = 29.99 years

Answer 4

I got \(A = $2951.52\) for the first part. However, for the second part, if it takes 30 years to get to $2951.52, doesn't 29.99 years seem like a long time to get to $1200*2=$2400? That's a heckuva interest buildup in the last 3.6 days :-)

Answer 5

Sorry, not $2400, $2000, argument still applies :-)

Answer 6

You probably didn't use \(A = $2000\) for the second part, I'm guessing.

Answer 7

ln (2951/1200)/.03 is that not right?

Answer 8

No, you're trying to find the time at which the balance (A) has increased to $2000. I used the same letter because it's the same formula, but it has a different value in the second part of the problem, right?

Answer 9

To spell it out, second part is to find the time \(t\) at which \(A = $2000\):
\[A = Pe^{rt}\]\[A = $2000\]\[P=$1200\]\[r=0.03\]\[\frac{\ln\frac{A}{P}}{r}=t = \frac{\ln\frac{2000}{1200}}{0.03} = \]

Answer 10

Here's a graph showing the point at which we get to the magic $2k. Note that the axes are a bit different than you might expect, as the x-axis is at $1200 here, so the bottom left corner of the graph is the starting point of $1200.