How long until $1000 invested at 6% compounded monthly grows to $2000? Step by Step

Step by step

The question

Question: How long until $1000 invested at 6% compounded monthly grows to $2000?

1 Write the equation

The compound interest rule, with r/n = 0.06 divided by 12: 0.005.

A = P(1 + rn)^nt, 0.06 ÷ 12 = 0.005

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

t is the unknown, up in the power.

1000 × 1.005^12t = 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~~ × 1.005^12t~~1000~~ = 20001000

1000 divided by 1000 is 1.

So just the power is left.

2000 divided by 1000 is 2.

Now we have 1.005 to the power 12t = 2.

1.005^12t = 2

Careful!

When the power comes down, all of it comes down: 12t, not just t.

Careful: the whole power 12t comes down, not only t

3 Take the log of both sides

Logs bring a power down.

So we take the log of both sides.

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

log(1.005^12t) = log 2

4 Bring the power down

The power rule for logs: the power comes down in front, as times.

log(aⁿ) = n × log a

The whole power, 12t, comes down in front.

12t × log 1.005 = log 2

5 Get t alone

Divide both sides by 12 times log 1.005.

So t is alone.

t = log 212 × log 1.005

6 Use a calculator

With a calculator: log 2 is about 0.30103.

And log 1.005 is about 0.00216606.

log 2 ≈ 0.30103

log 1.005 ≈ 0.00216606

So t is about 11.58 years, to two decimal places.

t ≈ 11.58

7 Check

Let's check: put t = 11.58 back in.

With a calculator.

We get about $1999.84.

1000 × 1.005^{12 × 11.58} ≈ 1999.84

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 the log of both sides.

Then: bring the power down.

Then: get t alone.

Then: use a calculator.

Last: check.

Answer