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A substance has a half-life of 10 days. How long until 60 mg decays…

Step by step, with the reason for every step · Half-life and time with logs · Algebra 2

Step by step

The question

Question: A substance has a half-life of 10 days.

How long until 60 mg decays to 20 mg?

1 Write the equation

The amount left is the start, times (1/2) to the power t/h: one halving per half-life.

A = A0 × (12)th

We know the start, 60 mg, and the end, 20 mg.

t is the unknown, up in the power.

60 × (12)t10 = 20

2 Get the power alone

To remove times 60, we divide by 60.

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.

60 × (12)t1060 = 2060

60 divided by 60 is 1.

So just the power is left.

20 divided by 60 is 1/3.

Now we have (1/2) to the power t/10 = 1/3.

(12)t10 = 13

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((12)t10) = log (13)

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, t/10, comes down in front.

t10 × log (12) = log (13)

5 Get t alone

Divide both sides by log (1/2), then times 10, to get t alone.

t = 10 × log (1/3)log (12)

6 Use a calculator

With a calculator: log 1/3 is about −0.477121.

And log (1/2) is about −0.30103.

log (13) ≈ −0.477121
log (12) ≈ −0.30103

So t is about 15.85 days, to two decimal places.

t ≈ 15.85

7 Check

Let's check: put t = 15.85 back in.

With a calculator.

We get about 20 mg.

60 × (12)15.85 ÷ 10 ≈ 20

That is 20, up to the rounding.

It works!

≈ 20

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
t ≈ 15.85

Your turn

Try this one yourself, then check:

Show the answer
t ≈ 3.47 days

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