Question

What interest rate compounded daily(365days/year) is required for a $22000 investment to grow to $38500 in 5 years ?
The amount $22,000 will grow to $38500 if the interval rate would be ___%
@ganeshie8
@mathmale
@mathstudent55

Answer 1

@ganeshie8
@mathmale
@mathstudent55

Answer 2

may need to use the formula for compounding but continuously

Answer 3

For the first part you need to solve r in

\[ 38500= 22000*(1+r/(365*100))^{365*5}\]

Answer 4

I thought I'd make a graphic to make that equation a little neater.

In order to solve the equation for "r", we must take the logarithm of both sides.

(See attachment)

How do you take the logarithm of (1 +r/36,500) ^ 1,825 ?

Answer 5

Okay, I have calculated that formula a little bit more and it is attached.
So, I guess the question now becomes
what is the logarithm of (1 +r/36,500) ?

Answer 6

Okay, let's try this:
Attached is a formula from this page:
http://www.1728.org/compint3.htm

Using the formula
exp = log(total / principal) / (compounding periods * years)
exp = log(38,500 / 22,000) / (365 * 5)
exp = 0.243038048686294 / 1,825
exp = 0.000133171533526737

Now, we need to input that number into this formula:
rate = (10^exp -1) * (compounding periods)
rate = (10^0.000133171533526737 -1) * (365)
rate = (1.00030668580639 -1) * (365)
rate = (.00030668580639) * 365
rate = 0.1119403193
rate = 11.194%
Which is the daily rate

To be extra sure, the calculator here:
http://www.1728.org/compint.htm
calculates the daily interest rate to be
11.194%

Yes, it's just that simple :-)