At what interest rate would you need to invest $80,000 in order to have $100,000 in 10 years, compounded continuously.

Question

Answer the question by completing each box below.

Box 1 - Information

Box 2 - Find

Box 3 - Formula

Box 4 - Plug & Chug

Box 5 - Units

Answer 1

The underlying formula that you want to use here for continuous compounding is

A = P*e^(rt), which when solved for the interest rate (r) becomes

A/P = e^(rt)

ln(A/P) = rt

r = ln(A/P) / t

for this scenario then that formula becomes

r = ln(100000/80000) / 10

r = ln(1.25) / 10

r = 0.022314 = 2.23%

meaning that to grow your $80,000 investment to $100,000 in 10 years with continuous compounding, you would need an annual interest rate of approximately 2.23%

Answer 2

Formula is A = Pe^rt where A is target amount, P is initial investment, e is the constant 2.71828, r is the interest rate as a decimal and t is number of compounding periods.

I'll do everything in $000's. A = $100; P = $80 and t = 10 yrs.

100 = 80e^10r

Divide both sides by 80: 1.25 = e^10r

Take natural log (ln) of both sides: ln1.25 = 10r

0.2231 = 10r; r = 0.02231, or 2.23%