Year 11: How to solve application of logarithm word problems

In Year 11 Mathematics in Australia, logarithms are used to solve exponential equations, especially when the unknown is in the exponent. Typical situations include:

How long it takes for an investment to reach a certain value
How long until a population reaches a target size
How long until a decaying substance falls to a certain amount

These questions usually start with a real-life situation, then lead to an exponential equation like $A = P \times {(1 + r)}^{t}$ . Logarithms help you “bring down” the exponent so you can solve for $t$ or for the rate $r$ .

Overall strategy

Understand the problem and build the exponential equation.
Rearrange to isolate the exponential part.
Take logarithms of both sides and solve for the unknown.
Interpret and check your answer.

Stage 1: understand and build the exponential model

Step 1: identify key quantities

As you read the word problem, identify:

The initial amount (for example, starting money, initial population, starting mass).
The growth or decay rate (usually a percentage).
The time period for that rate (per year, per month, etc.).
The final amount or the target amount.
What you are asked to find (often the time).

Step 2: define variables and write the exponential equation

Use clear variables, for example:

$A$ for the amount after time $t$
$P$ (or $A_{0}$ ) for the initial amount
$t$ for time
$r$ for the rate as a decimal (for example, 5% becomes 0.05)

The most common model is the compound interest or percentage change formula:

$A = P \times {(1 + r)}^{t}$ (growth)
$A = P \times {(1 - r)}^{t}$ (decay)

For half-life problems, where a quantity halves every $h$ time units:

$A = A_{0} \times {(1 / 2)}^{t / h}$

Stage 2: rearrange to use logarithms

Step 3: isolate the exponential part

Start from the exponential model and rearrange to get something like:

$k = b^{t}$

Here $k$ and $b$ are known numbers, and $t$ is the unknown.

Example with compound interest:

$A = P \times {(1 + r)}^{t}$
Divide both sides by $P$ :

$A / P = {(1 + r)}^{t}$

Step 4: take logarithms of both sides

Take logarithms of both sides. You can use $\log$ (base 10) or $\ln$ (natural logarithm). Use the same on both sides.

For example, starting from $A / P = {(1 + r)}^{t}$ :

$\log (A / P) = \log ({(1 + r)}^{t})$

Now use the log power rule:

$\log (b^{t}) = t \times \log (b)$

Applying this to the right-hand side:

$\log (A / P) = t \times \log (1 + r)$

Stage 3: solve for the unknown

Step 5: solve for time when the rate is known

From $\log (A / P) = t \times \log (1 + r)$ , solve for $t$ :

$t = \log (A / P) \div \log (1 + r)$

In general, if $k = b^{t}$ , then:

$t = \log (k) \div \log (b)$

Step 6: solve for the rate when time is known

Sometimes, you know $A$ , $P$ and $t$ , and you are asked to find the rate $r$ . Start from:

$A = P \times {(1 + r)}^{t}$

Divide both sides by $P$ :

$A / P = {(1 + r)}^{t}$

Take logs:

$\log (A / P) = t \times \log (1 + r)$

Divide both sides by $t$ :

$\log (A / P) \div t = \log (1 + r)$

Now undo the logarithm to solve for $1 + r$ . For example, using base 10 logs:

$1 + r = 10^{\log (A / P) \div t}$
finally, $r = 10^{\log (A / P) \div t} - 1$ .

In practice, you may prefer to use a scientific calculator’s “antilog” or simply use the $10^{x}$ or $e^{x}$ button depending on which type of log you used.

Worked example: investment doubling time

The problem

A savings account pays 5% interest per year, compounded annually. You invest 2000 dollars. How many years will it take for your investment to double?

Step 1: identify and model

Initial amount: $P = 2000$ dollars
Final amount: $A = 4000$ dollars (double)
Rate: $r = 0.05$
Unknown: time $t$ in years

Use the compound interest model:

$A = P \times {(1 + r)}^{t}$

Substitute values:

$4000 = 2000 \times {(1.05)}^{t}$

Step 2: isolate the exponential part

Divide both sides by 2000:

$4000 \div 2000 = {(1.05)}^{t}$
$2 = {(1.05)}^{t}$

Step 3: take logarithms

Take base 10 logs of both sides (you could also use natural logs):

$\log (2) = \log ({(1.05)}^{t})$

Use the power rule:

$\log (2) = t \times \log (1.05)$

Step 4: solve for t

Rearrange:

$t = \log (2) \div \log (1.05)$

Using a calculator (values are approximate):

$\log (2) \approx 0.3010$
$\log (1.05) \approx 0.0212$

So:

$t \approx 0.3010 \div 0.0212 \approx 14.2$ years

It will take about 14.2 years for the investment to double at 5% annual interest.

Step 5: reasonableness check

A common rough rule is “doubling time is approximately 70 divided by the percentage rate”. Here, 70 ÷ 5 = 14 years, which matches our answer well.

Outline example: half-life

The problem

A radioactive substance has a half-life of 10 years. You start with 200 grams. After how many years will only 25 grams remain?

Model:
Half-life model: $A = A_{0} \times {(1 / 2)}^{t / h}$
Here $A_{0} = 200$ , $A = 25$ , $h = 10$ .
Substitute values:
$25 = 200 \times {(1 / 2)}^{t / 10}$
Isolate exponential part:
Divide both sides by 200:
$25 \div 200 = {(1 / 2)}^{t / 10}$
$1 / 8 = {(1 / 2)}^{t / 10}$
Take logs and solve:
$\log (1 / 8) = (t / 10) \times \log (1 / 2)$
$t / 10 = \log (1 / 8) \div \log (1 / 2)$
$t = 10 \times (\log (1 / 8) \div \log (1 / 2))$
Since 1/8 is (1/2)³, you can also see that $t / 10 = 3$ , so $t = 30$ years.

It takes 30 years for 200 grams to decay to 25 grams.

Common mistakes and tips

Forgetting to divide by the initial amount first.
Always isolate the exponential part before taking logs.
Using the percentage instead of the growth factor.
For a 7% increase, the growth factor is 1.07, not 0.07.
Mixing bases of logarithms.
It is fine to use either $\log$ or $\ln$ , but stick to one base within a calculation.
Not including units.
Time must have units (years, months, hours). Amounts should have units (dollars, grams, people).

Summary checklist

Did you start from a correct exponential model?
Did you isolate the exponential expression before taking logs?
Did you use the log power rule correctly?
Did you solve clearly for the unknown and interpret the result with units?
Does your answer make sense in the real-world context?

If you follow this structure each time, logarithm word problems become much more systematic. With practice, you will be able to set up and solve these equations quickly and accurately.