Suppose $7,500 is compounded weekly for 26 years. If no other deposits are made, what rate is needed for the balance to double in that time? Round the answer to

Question

Suppose $7,500 is compounded weekly for 26 years. If no other deposits are made, what rate is needed for the balance to double in that time?
Round the answer to the nearest hundredth of a percent.

Answer 1

We'll need the compound interest formula for this one. 

 

A=P(1+r/n)^(nt), where A is the amount at the end, P is the principal (the initial amount you borrow or invest), r is the rate, n is the number of times the interest is compounded each year, and t is how many years the money is deposited for.

 

From the information given by the problem, we know that:

P=7,500

t=26

A=15000 (because that's double 7,500)

n=52 (because it's compounded weekly and there are 52 weeks in a year)

We don't know r, but now that's the only variable and we can set up our equation and solve.

 

15000=7500(1+r/52)^(52*26)

 

Simplifying, we get:    15000=7500(1+r/52)^1352

 

Divide both sides by 7500 and we get:     2=(1+r/52)^1352

 

Take the log of both sides:     log(2)=log(1+r/52)^1352

 

Using properties of logs, bring the 1352 in front:     log(2)=1352*log(1+r/52)

 

Divide both sides by 1352:    0.000226553222=log(1+r/52)

 

Switch back to exponential form:    10^(0.000226553222)=1+r/52

 

Subtract 1 from both sides:     0.00052179364=r/52

 

Multiply both sides by 52:    r=0.027133

 

As a percentage, this is 2.7%

 

 

 

There are many other ways to do the algebra, some of which may be easier. So if you have a different idea when solving for r, go for it!