What interest rate would you need to double $3500 in 9 years

Question

interest rate

Answer 1

Assuming annual compounding:

 

P = P₀(1 + r)^t

• P = amount after t years = 2($3500) = $7000
• P₀ = initial deposit = $3500
• r = interest rate expressed as a decimal
• t = years = 9

 

P = P₀(1 + r)^t

7000 = 3500(1+r)⁹

2 = (1+r)⁹

2^1/9 = 1 + r      [2^1/9 = ⁹√2 = ninth root of 2]

2^1/9 - 1 = r

 

Use a calculator to compute the answer.  r will be in decimal form.  Multiply by 100 to convert to percent.