Solve real exponential and logarithmic applications

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Choose a model, solve for the unknown, and explain what the result means

Exponential models describe repeated percentage change. They can represent growth, such as money earning interest, or decay, such as a substance losing part of its amount. In many applications, the starting amount and growth rate are known, but the time is not. A logarithm helps solve for that time. This lesson reviews the needed equation skills, builds models from context, and shows how to interpret a solution in the real situation.

What you will learn

Recognize situations where a quantity changes by a constant percentage over equal time periods.
Build an exponential model from an initial amount and a growth or decay factor.
Use logarithms to solve an exponential equation when the unknown is in the exponent.
Check that a solution is reasonable and state it with appropriate units.

1. Prerequisite bridge: amount, rate, and factor

An initial amount is the quantity at the beginning of a situation. A rate is a comparison that describes how a quantity changes. A percentage rate must be written as a decimal before it is used in a calculation: for example, a rate of 4% is

0.04

.

For a single increase of 4%, the new amount is the original amount plus 4% of it. That is the same as multiplying the original amount by

1.04

. For a decrease of 4%, the remaining amount is 96% of the original, so the multiplier is

0.96

.

An exponential change repeats the same multiplier over equal periods. If the amount is multiplied by

1.04

each year, after two years it has been multiplied twice. The exponent counts how many periods of change have passed.

A=P(1+r)^n

Convert a percentage to a decimal before calculating.
Growth factor: $1+r$ . Decay factor: $1-r$ , where $r$ is the rate as a decimal.
The exponent counts equal time periods, so its unit must match the rate period.

2. Build and read an exponential model

In the model,

P

is the initial amount,

A

is the amount after the change,

r

is the rate per period, and

n

is the number of periods. The factor

1+r

describes growth. For decay, use

1-r

instead. The model assumes the same percentage change occurs each period.

A concrete example is a savings balance that grows by 4% each year. If the starting balance is CAD 500, the amounts after zero, one, and two years are CAD 500, CAD 520, and CAD 540.80. The increases are not the same number of dollars, because each increase is calculated from the current balance. The multiplier remains the same.

A table makes the repeated factor visible. A graph of the same situation would plot time on the horizontal axis and amount on the vertical axis. For growth, the points rise and the amount increases more quickly over time. For decay, the points fall toward smaller positive amounts. The model, table, and graph are different ways to describe the same relationship.

A=P(1-r)^n

Use the starting amount as the value at period zero.
A constant percentage change does not usually produce a constant dollar change.
Check whether the context describes growth or decay before choosing the factor.

3. Use logarithms to find an unknown time

A logarithm answers the question: to what exponent must a base be raised to produce a given value? It is the inverse operation of exponentiation. For example, if an equation has an unknown in its exponent, a logarithm can bring that exponent down so it can be found.

First isolate the exponential expression. Then take the logarithm of both sides. The same logarithm must be applied to both sides, so the equality is preserved. Use a calculator's natural logarithm button, , or common logarithm button, , consistently. The quotient rule for logarithms gives the unknown exponent.

A calculated time may not be a whole number of periods. In a context where changes happen only at the end of each period, interpret the result carefully. If the question asks when a target is first reached, round up to the next whole period when needed. Also check that the answer makes sense: a growth target above the initial amount should require positive time.

(b^x)=x(b)

A logarithm is used to solve when the unknown is an exponent.
Apply the same logarithm to both sides of the equation.
Interpret time using the period in the situation, and include units.

4. Apply, check, and communicate

Start an application by naming the unknown and identifying the known information. Decide what one period means: a year, a month, or another interval. Match the rate to that period. For example, a monthly rate needs a number of months as its exponent.

After solving, substitute the result into the original model or use the model to estimate the amount at a nearby whole period. This checks both the arithmetic and the interpretation. In a word problem, report a sentence that answers the question, rather than only writing a number.

A model is an approximation of a real situation. Its assumptions matter. A fixed percentage rate may describe a balance under a stated interest plan, but an actual balance could also depend on fees, deposits, or changing rates. Use only the conditions given in the problem.

n=\frac{\log(A/P)}{\log(1+r)}

Define the unknown and state its unit before solving.
Keep the time unit consistent with the rate period.
Check the result against the context and report it in a complete sentence.

First periods of a balance growing by 4% per year

Years	Balance (CAD)	How it is found
0	500.00	Starting amount
1	520.00	500.00 × 1.04
2	540.80	520.00 × 1.04

Worked example

Finding when a savings target is reached

A savings account starts with CAD 1,200 and earns 4.2% annual interest, compounded monthly. Assuming no deposits or withdrawals, after how many months will the balance first be at least CAD 1,600?

Identify the period and factor
The rate is annual, but interest is compounded monthly. Divide the annual rate by 12 to get the rate per month. The account grows, so add the monthly rate to 1. Let $n$ be the number of months.
$1+\frac{0.042}{12}=1.0035$
Write the balance model
The starting balance is CAD 1,200. Multiply it by the monthly growth factor once for each month. Set the balance equal to CAD 1,600 to find the time when the target is reached.
$1600=1200(1.0035)^n$
Isolate the exponential expression
Divide both sides by 1,200. This leaves the growth factor raised to the unknown number of months.
$(1.0035)^n=\frac{4}{3}$
Solve for the exponent
Take the natural logarithm of both sides. The logarithm power rule lets the exponent become a factor. Divide by the logarithm of the monthly growth factor to find $n$ .
$n=\frac{\ln(4/3)}{\ln(1.0035)}\approx82.33$
Interpret the time
The calculation gives about 82.33 months. With monthly compounding, the balance is evaluated at whole-month intervals. The target is therefore first reached at the next monthly period, month 83.
$\lceil82.33\rceil=83$

Answer: The balance first reaches at least CAD 1,600 after 83 months, or 6 years and 11 months.

Check: At month 82 the balance is about CAD 1,598.13, still below the target. At month 83 it is about CAD 1,603.72, which is above the target.

Common mistakes and how to avoid them

Using 4.2 as the rate instead of converting 4.2% to a decimal.

Correction: Write 4.2% as 0.042, then divide by 12 for the monthly rate in a monthly-compounding model.

Using the annual rate as the monthly rate when interest is compounded monthly.

Correction: Match the rate and exponent to the same period. For an annual rate compounded monthly, use the annual rate divided by 12 and count months.

Rounding a time down when the question asks when a target is first reached.

Correction: Check the amounts at the neighbouring whole periods. If the earlier period is below the target, the next period is the first one that reaches it.

Giving a calculated number without units or a contextual answer.

Correction: State what the number measures, such as months, years, or dollars, and answer the question in a sentence.

Lesson summary

Repeated percentage change can be modelled with an exponential equation.
Use a growth factor for increase and a decay factor for decrease.
When the unknown is in the exponent, use logarithms to solve for it.
Check the result, respect the time periods in the context, and state a meaningful answer.

Check your understanding

Question 1

A quantity starts at 800 and decreases by 15% each month. Which model gives the amount after

m

months?

$A=800(1.15)^m$
$A=800(0.85)^m$
$A=800-0.15m$
$A=800(0.15)^m$

Show answer and explanation

A=800(0.85)^m

A 15% decrease leaves 85% of the amount each month, so the repeated factor is 0.85.

Question 2

A model is

A=300(1.02)^n

. Which operation is useful for solving for

n

when

A

is known?

Subtract 300 from both sides
Divide both sides by 1.02
Take a logarithm of both sides after dividing by 300
Multiply both sides by 1.02

Show answer and explanation

Take a logarithm of both sides after dividing by 300

First isolate the power by dividing by 300. A logarithm then allows the exponent to be found.

Question 3

A balance reaches a target after a calculated 5.4 monthly periods. If it changes only at the end of each month, when is the target first reached?

After 5 months
After 5.4 months exactly
After 6 months
After 4 months

Show answer and explanation

After 6 months

The balance is evaluated at whole-month intervals. Five periods are not enough, so the first monthly interval at or beyond 5.4 is month 6.

Key terms

Initial amount: The quantity at the beginning of the situation.
Growth factor: The multiplier greater than 1 used for a repeated percentage increase.
Decay factor: The multiplier between 0 and 1 used for a repeated percentage decrease.
Exponential model: An equation in which a fixed factor is raised to a power that counts periods of change.
Logarithm: The exponent needed to raise a base to produce a specified value.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A3.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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