$1000 is invested at 5% compounded continuously. How long until it… Step by Step

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The question

Question: $1000 is invested at 5% compounded continuously.

How long until it doubles?

1 Write the equation

Continuous interest: A = P times e to the power rt.

A = Pe^rt

We know the start, $1000, and the end, $2000.

t is the unknown, up in the power.

1000 × e^0.05t = 2000

2 Get the power alone

To remove times 1000, we divide by 1000.

We do it on both sides.

Why both sides?

The = sign means both sides are the same amount.

If we change only one side, they are not the same any more.

~~1000~~ × e^0.05t~~1000~~ = 20001000

1000 divided by 1000 is 1.

So just the power is left.

2000 divided by 1000 is 2.

Now we have e to the power 0.05t = 2.

e^0.05t = 2

Careful!

When the power comes down, all of it comes down: 0.05t, not just t.

Careful: the whole power 0.05t comes down, not only t

3 Take ln of both sides

Logs bring a power down.

With e, we use ln.

So we take ln of both sides.

We do it on both sides, so they stay equal.

ln(e^0.05t) = ln 2

4 Bring the power down

ln and e undo each other: ln of e to the power k is just k.

ln(eᵏ) = k

So the left side is just 0.05t.

0.05t = ln 2

5 Get t alone

To remove times 0.05, we divide by 0.05.

We do it on both sides, so they stay equal.

t = ln 20.05

6 Use a calculator

With a calculator: ln 2 is about 0.693147.

ln 2 ≈ 0.693147

So t is about 13.86 years, to two decimal places.

t ≈ 13.86

7 Check

Let's check: put t = 13.86 back in.

With a calculator.

We get about $1999.71.

1000 × e^{0.05 × 13.86} ≈ 1999.71

That is 2000, up to the rounding.

It works!

≈ 2000

Let's look back.

First: write the equation.

Then: get the power alone.

Then: take ln of both sides.

Then: bring the power down.

Then: get t alone.

Then: use a calculator.

Last: check.

Answer