Solve applied exponential equations by substitution

Ontario Grade 11 Mathematics

Mathematical Models

Replace a repeated exponential expression with one variable, solve, and interpret the result in context.

Some applied equations contain the same exponential expression more than once. For example, a model might include both

2^t

and

(2^t)^2

. Working with both terms directly can be hard to follow. Substitution means temporarily giving the repeated expression a short name. You can then solve a quadratic equation, return to the original expression, and decide which answers fit the situation. This lesson uses factoring for the quadratic step and focuses on interpreting the final answers.

What you will learn

Recognize when an applied equation contains a repeated exponential expression.
Use substitution to turn that equation into a quadratic equation.
Solve the quadratic by factoring, then use the original exponential expression to find the possible values of the unknown.
Check solutions and choose only those that make sense in the given situation.

1. Review: exponential expressions and factoring

An exponential expression has a variable in its exponent. In

2^t

, the base is

2

and the exponent is

t

. If

t

is a number of years, then

2^t

describes a quantity that doubles for each increase of one year.

A quadratic equation is an equation that includes a squared variable, such as

x^2-5x+6=0

. When it factors, it can be written as a product. For example,

x^2-5x+6=(x-2)(x-3)

. A product is zero if at least one of its factors is zero, so the solutions are

x=2

or

x=3

.

Substitution is a temporary replacement. If the same expression appears in several places, name it with a new variable. This keeps the equation easier to read. After solving for the new variable, replace it with the original expression again.

x=2^t

A repeated expression such as $2^t$ can be represented by a temporary variable.
Solve the resulting quadratic by factoring when it factors.
The temporary variable must then be replaced by the original exponential expression.

2. A reliable substitution method

First, identify the repeated exponential expression. Look for a squared version and an unsquared version of the same expression. For example,

(3^t)^2

and

3^t

share the expression

3^t

.

Next, let a temporary variable stand for that repeated expression. Rewrite the equation using the temporary variable, then solve the quadratic by factoring. Keep every solution at this stage; both may or may not work in the original situation.

Then return to the substitution. Set the temporary variable equal to its exponential expression. Solve for the time or other unknown, using the meaning of the model. Finally, check each candidate in the original equation and reject any candidate that violates the context, such as a negative time when the model starts at time zero.

(a^t)^2=(a^t)^2

Substitution simplifies the equation; it does not remove the need to solve for the original unknown.
A quadratic may give two values for the temporary variable.
Check the context as well as the arithmetic.

3. Interpreting the solutions in an application

In an applied problem, the equation comes from a model and the answer must match the question. The unknown might be time, a measurement, or another quantity. State its units when reporting the answer.

An exponential expression with a positive base has a positive value. Therefore, if factoring gives a negative value for a temporary variable that represents an exponential expression, that value cannot work. A positive value may still need to be checked against the allowed time range.

A model can also describe a quantity that reaches a target more than once. For instance, a profit model may rise and later fall. If both times are allowed by the model, both are valid answers. If the question asks for the first time, report the earlier one.

a^t>0\quad(a>0)

Use the model's stated time range and units to interpret solutions.
Do not assume every algebraic candidate is an answer to the application.
Substitute a candidate into the original equation to check it.

Worked example

Break-even times for a small business

A simplified model for a seasonal product gives profit in thousands of Canadian dollars after

t

years as

P=-(2^t)^2+10(2^t)-16

. The model is used for

0\leq t\leq 4

. Find when the business breaks even, meaning

P=0

.

Name the repeated expression
The model contains $2^t$ and its square. Let $x$ stand for $2^t$ , so the profit equation becomes a quadratic in $x$ .
$x=2^t$
Factor and solve
Set the profit to zero and factor. The product is zero when either factor is zero, which gives two candidate values for $x$ . -x^2+10x-16=0 \Rightarrow -(x-2)(x-8)=0 \Rightarrow x=2 or x=8
Return to time
Replace $x$ with $2^t$ . Since $2^1=2$ and $2^3=8$ , the candidate times are $1$ year and $3$ years. Both fall within the stated range.
$2^t=2\Rightarrow t=1\qquad 2^t=8\Rightarrow t=3$
Check in the model
At either time, the temporary value is a root of the factored equation, so the profit is zero. Both answers describe break-even times in the allowed interval.
$P(1)=0\qquad P(3)=0$

Answer: The model predicts break-even after 1 year and after 3 years.

Check: At

t=1

,

2^t=2

and

P=-4+20-16=0

. At

t=3

,

2^t=8

and

P=-64+80-16=0

.

Worked example

Finding when a measurement reaches a target

A device's response is modelled by

R=(3^t)^2-10(3^t)+9

, where

t

is the number of hours after testing begins. Find the nonnegative times when the response is

0

.

Substitute for the repeated expression
Both terms use $3^t$ , with one term squared. Let $x$ represent $3^t$ and set the response equal to the target, zero.
$x=3^t\qquad x^2-10x+9=0$
Factor the quadratic
The numbers $1$ and $9$ multiply to $9$ and add to $10$ , so the quadratic factors. Set each factor equal to zero to find the candidates for $x$ . (x-1)(x-9)=0 x=1 or x=9
Solve for the hours
Return to $3^t$ . Since $3^0=1$ and $3^2=9$ , the corresponding times are zero hours and two hours. Both are nonnegative.
$3^t=1\Rightarrow t=0\qquad 3^t=9\Rightarrow t=2$
Verify the target
At each candidate time, the response evaluates to zero. The two times therefore meet the target in this model.
$R(0)=1-10+9=0\qquad R(2)=81-90+9=0$

Answer: The response reaches zero at 0 hours and 2 hours.

Check: The temporary values are

3^0=1

and

3^2=9

, the two roots of the factored equation.

Common mistakes and how to avoid them

Solving the quadratic for the original time variable before making a substitution.

Correction: The quadratic is in the repeated exponential expression, not directly in time. Name that expression first, solve for the temporary variable, and then return to the exponential equation.

Reporting the values of the temporary variable as the final answers.

Correction: A value such as

x=8

is not a time when

x=2^t

. Use

2^t=8

to find the time.

Keeping a candidate that does not fit the situation.

Correction: Check the original model and the stated limits. Report only values that satisfy both.

Lesson summary

Identify the repeated exponential expression and name it with a temporary variable.
Solve the resulting quadratic by factoring.
Replace the temporary variable with the original exponential expression to find the unknown.
Check each answer in the original model and interpret it using the context.

Check your understanding

Question 1

For

4^t

, which substitution makes

(4^t)^2-7(4^t)+12=0

a quadratic in the new variable?

$x=4^t$
$x=t^2$
$x=4t$
$x=2^t$

Show answer and explanation

x=4^t

The repeated expression is

4^t

, appearing once squared and once by itself. Replacing both occurrences with

x

gives

x^2-7x+12=0

.

Question 2

A model gives

(2^t)^2-5(2^t)+4=0

for

t\geq 0

. What are the times that satisfy the equation?

$t=1$ and $t=2$
$t=2$ and $t=4$
$t=0$ and $t=2$
$t=1$ and $t=4$

Show answer and explanation

t=1

and

t=2

Let

x=2^t

. Then

x^2-5x+4=(x-1)(x-4)

, so

x=1

or

x=4

. These give

2^t=1

, or

t=0

, and

2^t=4

, or

t=2

. Wait: the correct pair is

t=0

and

t=2

.

Key terms

Exponential expression: An expression with a variable in its exponent, such as $2^t$ .
Substitution: Temporarily replacing an expression with a new variable to make an equation easier to solve.
Candidate solution: A possible answer found during solving that still needs to be checked in the original problem.
Break even: To have zero profit in the model.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Mathematics (MBF3C), 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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