Please help with this word problem. Thanks!

[math-a-phobic]

Hi and thank you for your help in advance. I tried to solve this problem but got stuck.

Problem: You deposit $1000 into each of two savings account. One with an annual percentage rate of 8.05% compounded monthly and antoehr with an annual percentage rate of 8% compounded continuously. How long would it take for the balance in one of the accounts to exceed the balance in the other account by $100? By $100,000?

This was as far as I got. I got $1083.54 for the monthly compounded account with a principal of $1000. For the other account, I got $1083.29. I hope that I at least got that part right. :? I don't even know where to begin to find the solution to the questions about the year when the balance starts to exceed.

 

[galactus]

Here's one approach:

I assume you know the formulae for compunding interest?.

continuously: \(\displaystyle \L\\A=Pe^{rt}\)

monthly: \(\displaystyle \L\\A=P(1+\frac{r}{n})^{nt}\)

So, you have for your problem:

\(\displaystyle \L\\1000e^{.08t}\)

\(\displaystyle \L\\1000(1+\frac{.0805}{12})^{12t}\)

Now, the difference must be 100 dollars:

\(\displaystyle \L\\1000e^{.08t}-1000(1+\frac{.0805}{12})^{12t}=100\)

Solve for t. Do the same for the 100,000.

 

[math-a-phobic]

I'm sorry for asking for your help again, I tried to solve for t but didn't know how to. Should I divide by LN .08 since it is the opposite of e^(.08t) and this will let me get t to be alone?

 

[galactus]

Don't be sorry. I just hope I can help.

Actually, this is a booger of a problem to solve for t. Maybe one of the

learned contributors has a more efficient method.

\(\displaystyle \L\\1000(\frac{24161}{24000})^{12t}-1000e^{\frac{2}{25}t}=100\)

\(\displaystyle \L\\1000(\frac{24161}{24000})^{12t}-e^{\frac{2}{25}t})=100\)

\(\displaystyle \L\\(\frac{24161}{24000})^{12t}-e^{\frac{2}{25}t}=\frac{1}{10}\)

\(\displaystyle \L\\ln(\frac{24161}{24000})^{12t}=ln(\frac{1}{10}+e^{\frac{2}{25}t})\)

\(\displaystyle \L\\12tln(\frac{24161}{24000})=ln(\frac{1}{10}+e^{\frac{2}{25}t})\)

\(\displaystyle \L\\12ln(\frac{24161}{24000})t=ln(\frac{1}{10}+e^{\frac{2}{25}t})\)

\(\displaystyle \L\\e^{12(\frac{24161}{24000})t}=\frac{1}{10}+e^{\frac{2}{25}t}\)

\(\displaystyle \L\\e^{12ln(\frac{24161}{24000})t}-e^{\frac{2}{25}t}-\frac{1}{10}=0\)

Now, use Newton's method with the initial guess set at 30:

Like I said, this is one feeble attempt but it gives you the answer. Perhaps someone else has a better method. You could also try graphing the two and seeing where they intersect. The graphs are virtually identical, so be prepared to zoom. Graph \(\displaystyle \frac{1}{10}+e^{.08x}\) and
\(\displaystyle (\frac{24161}{24000})^{12x}\)

Frankly, this is where our nice modern technology comes in handy.