Question 43603
A = P(1 + r/n)^m.

Suppose you deposit $10,000 for 2 years at a rate of 10%.  Calculate the return (A) if the bank compounds annual (n=1).

A=10000(1+0.10/1)^2
A=10000*1.21= 12100

Calculate the return (A) if the bank compounds quarterly (n = 4), and carry all calculations to 7 significant figures. (I assume you mean for 2 years)
A=10000(1+0.10/4)^(4*2)
A=10000(1.2184029
A=12184.03

Calculate the return (A) if the bank compounds monthly (n=12), and carry all calculations to 7 significant figures. (I assume you for 2 years) 
A=10000(1+0.10/12)^(12*2)
Use your caluculator.


Calculate the return (A) if the bank compounds daily (n=365), and carry all calculations to 7 significant figures. (I assume you mean for two years)

A=10000(1+0.10/365)^(365*2)
Use calculator.

What observation can you make about the increase in your return as your compounding increases more frequently?
That should be obvious.

If a bank compounds continuously, then the formula takes a simpler, that is A = Pe^n.  Where e is a constant and equals approximately 2.7183. Calculate A with continuous compounding.
You have to be given the rate and the number of years.

A=Pe^(rn)
A=10000(e^(0.10*?)

Now suppose, instead of knowing t, we know that the bank returned to us $15,000 with the bank compounding continuously.  Using natural logarithms, find how long we left the money in the bank (find f).

15000=10000e^(0.10t)
1.5=e^0.10t
ln(1.5)=0.10t
0.40546511...=0.10t
t=4.05 years

"How long will it take to double my money?" At 10% interest rate and continuous compounding

2P=Pe^(0.10t)
2=e^(0.10t)
ln2=0.10t
0.69314718...=0.10t
t=6.93 years
This is generally call "the rule of seven".

Cheers,
Stan H.