Exponential Equations 15

Question

Find the time it takes for $6,100 to double when invested at an annual interest rate of 10%, compounded continuously.

 years  

Find the time it takes for $610,000 to double when invested at an annual interest rate of 10%, compounded continuously.

 years  

Give your answers accurate to 4 decimal places.

Answer 1

Question: Find the time it takes for $6,100 to double when invested at an annual interest rate of 10%, compounded continuously. Find the time it takes for $610,000 to double when invested at an annual interest rate of 10%, compounded continuously. Give your answers accurate to 4 decimal places.

Answer: Money, money, money!

Since we have continuously compounding interest, the compounding periods are infinitesimally small, so we get to pull out the PERT formula! That is, A = Pe^rt. In the first part, we're looking to double our initial investment (principal) of $6,100, that is, to get $12,200 (plugged into A in the formula) at the end of the investing time. We also know we have a 10% interest rate (represented as 0.1 and plugged into r in the formula). It's our job to find the time it takes to double the principal, and there will be logarithms.

(12,200) = (6,100)e^(0.1)t is the set-up.

Divide both sides by 6,100 to simplify things a bit:

2 = e^0.1t.

Now we need to knock down the 0.1t from its pedestal. Take the natural log of both sides:

ln(2) = ln(e^0.1t). (Remember one of your natural log properties: ln(e^x) = x)

We then get ln(2) = 0.1t.

Divide both sides of the equation by 0.1 (or multiply both sides by 10, either way gets the same result):

t ≈ 6.9315.

It will take approximately 6.9315 years (or just slightly above 6 years and 11 months) for your initial investment of $6,100 to double in a 10% account compounded continuously.

The second part of the question applies exactly the same logic as in the first part, but with different numbers. ($610,000 principal, 10% interest, continuous compounding)

We are looking for $1,220,000 at the end of the investment ($610,000 x 2 = $1,220,000).

(1,220,000) = (610,000)e^(0.1)t. Divide both sides by 610,000:

2 = e^0.1t.

.......

This is the same as before! The idea here is that if you keep the interest rate at 10% for a bunch of problem like this where you seek to DOUBLE the money (to multiply the initial principal P₀ by 2), you end up with a scenario where you will get 2P₀ = P₀e^0.1t ---> 2 = e^0.1t. Feel free to change the interest rate to what you'd like, but keep the 2 where it is in the simpler equation, and you'll have a formula for the time it takes to double your money. Just take some natural logs and work it out! You'll also see patterns like this for archaeology, carbon-dating, half-life, and bacteria word problems.

.......

2 = e^0.1t. As before, t ≈ 6.9315.

It will take approximately 6.9315 years (or just slightly above 6 years and 11 months) for your initial investment of $610,000 to double in a 10% account compounded continuously.