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A1

Algebra 1 View details

Table of contents

Chapter 6

5.

Solving Exponential Equations

Solving exponential equations requires a deep understanding of mathematical principles. The lesson delves into the methods and techniques to tackle these equations, both graphically and algebraically. One of the key strategies is to rewrite the equations so that both sides have a common base, allowing for easier comparison and solution. The Property of Equality for Exponential Equations plays a pivotal role, stating that if two exponential expressions with the same base are equal, their exponents must also be equal. This principle is foundational in solving such equations. The lesson also presents various examples and scenarios to illustrate the application of these methods, aiding learners in grasping the concepts effectively.

Lesson Settings & Tools

	11 Theory slides
	7 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Solving Exponential Equations

Slide of 11

This lesson will define and discuss equations that involve exponential expressions.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

When Will the Cities Have the Same Population?

Dominika has a meeting with her guidance counselor on Monday afternoon to discuss her college plans. She is considering where she wants to apply in a few years. Dominika knows that she wants to study in a large city, but not one that is too big. She has two cities in mind. To decide between them, she is paying close attention to their populations, which are given by two exponential functions.

Functions f(x)=8(1.0048)^2 and g(x)=3.5(1.0067)^x that describe population of two cities

Here, $x$ is the number of years that have passed since the year $2000 .$ Furthermore, $f$ and $g$ are the populations of each city in millions of people after $x$ years. If they keep growing like this, in what year will the populations be the same? Approximate the answer to the nearest century.

Discussion

Exponential Equations

An exponential equation is an equation where variable expressions occur as exponents. As with any kind of equation, there are different types of exponential equations.

	Example Equation
With One Variable	$2^{x} = 32$
With the Same Variable on Both Sides	$2^{2 x} = 5 \cdot 2^{x}$
With the Same Base	$4^{3 x} = 4^{2 x + 3}$
With Unlike Bases	$3^{x + 4} = 8 1^{x}$
With a Rational Base	$(\frac{1}{2})^{x} = 8$

An exponential equation can be solved graphically.

Method

Solving Exponential Equations Graphically

An example exponential equation will be considered.

2^{x + 1} = 4^{x}

There are three steps to follow to solve exponential equations graphically.

1

Create Two Functions From the Given Equation

Each side of the equation can be considered as a function. In this case, both sides can be written as exponential functions.

2^{x + 1} = 4^{x} \Rightarrow y = 2^{x + 1} y = 4^{x} (I) (II)

The idea is that each function should be relatively easy to graph. In this case, the graph of the first function is a translation one unit to the left of the graph of its parent function

y = 2^{x} .

The second function is a parent function itself and its graph does not undergo any transformations.

2

Graph the Functions

Now, both functions will be graphed on the same coordinate plane.

The graphs of the functions y=2^(x+1) and y=4^x.

The number of solutions to the equation is the number of points of intersection of the graphs.

3

Identify the

x -

coordinates of the Points of Intersection

The solutions to the equation are the $x -$ coordinates of any points of intersection of the graphs. Since these graphs intersect at one point, the equation has one solution.

functions

The

x -

coordinate of the point of intersection is

1 .

Therefore, the solution to the equation is

x = 1 .

This can be verified by substituting

1

for

x

in the given equation and checking whether a true statement is obtained.

2^{x + 1} = 4^{x}

$x = 1$

2^{1 + 1} = ? 4^{1}

▼

Evaluate

Add terms

2^{2} = ? 4^{1}

Calculate power

4 = 4 ✓

Since a true statement was obtained,

x = 1

is a solution to the equation. Note that if the point of intersection is not a lattice point, the exact solution may not be easy to find using this method.

Example

Savings Account

Dominika's first class on Mondays is economics and personal planning. She is told that a certain savings account earns $6 %$ annual interest compounded yearly.

ATM

Suppose that Dominika deposits

$ 500

and wants to determine how many years it will take her to have

$ 800

in this account from interest alone. To do so, she must solve the following exponential equation.

500 (1.06)^{x} = 800

Help Dominika solve this equation! Round the answer to the nearest integer.

Hint

Consider each side of the equation as a function.

Solution

The given exponential equation will be solved graphically. To do this, each side will be considered as a function.

500 (1.06)^{x} = 800 ⇓ y = 500 (1.06)^{x} y = 800 (I) (II)

The first is an exponential function and the second a constant function. They will both be graphed on the same coordinate plane. The graph of the constant function is a horizontal line whose

y -

800 .

To graph the exponential function, the initial value

500

and the constant multiplier

1.06

will be used.

The graphs intersect at one point, so there is only one solution to the equation.

The $x -$ coordinate of the point of intersection appears to be $8 .$ However, looking closely at the graph, it can be seen that the $x -$ coordinate of this point is a bit greater than $8 .$

Fortunately, the answer should be rounded to the nearest integer. Therefore, the solution to the equation is

x \approx 8 .

This solution can be checked by substituting the value into the given equation.

500 (1.06)^{x} = 800

$x \approx 8$

500 (1.06)^{8} \approx ? 800

▼

Evaluate left-hand side

Calculate power

500 (1.593848 \dots) \approx ? 800

Multiply

796.924037 \dots \approx 800 ✓

Since a true statement was obtained, it is confirmed that the solution to the equation is

x \approx 8 .

This means that Dominika will have

$ 800

in her account in about

8

years.

Example

More Than One Solution

Dominika's second class on Mondays is math. As soon as she enters the classroom, she sees the following equation written on the board.

2 (3)^{x} = \frac{5}{3} x + \frac{7}{3}

The teacher says that the equation has two solutions and that when solved by graphing, the first solution can be exactly determined but that the second can only be approximated. Write the exact solution for Dominika.

Approximate the second solution to the nearest integer.

Hint

Consider each side of the equation as a function. Then graph the functions and find their point of intersection.

Solution

The equation will be solved graphically. To do so, each side of the equation will be considered as a function.

2 (3)^{x} = \frac{5}{3} x + \frac{7}{3} ⇓ y = 2 (3)^{x} y = \frac{5}{3} x + \frac{7}{3} (I) (II)

The first function is an exponential function and the second one is a linear function. Both can be graphed on the same coordinate plane. The slope and

y -

intercept of the linear function will be used to graph it. To graph the exponential function, its initial value and constant multiplier will be used.

The graphs intersect at two points. Therefore, the equation has two solutions, which are the $x -$ coordinates of these points of intersection.

It is seen that

x = - 1

is an exact solution. Conversely, the exact value of the second solution cannot be determined from the graph. However, approximated to the nearest integer, this solution is

x \approx 0 .

These solutions can be verified by substituting into the given equation. First,

x = - 1

will be checked.

2 (3)^{x} = \frac{5}{3} x + \frac{7}{3}

$x = - 1$

2 (3)^{- 1} = ? \frac{5}{3} (- 1) + \frac{7}{3}

▼

Evaluate

NegExponentToFrac

$a^{- m} = \frac{1}{a ^{m}}$

2 (\frac{1}{3 ^{1}}) = ? \frac{5}{3} (- 1) + \frac{7}{3}

$a^{1} = a$

2 (\frac{1}{3}) = ? \frac{5}{3} (- 1) + \frac{7}{3}

Multiply

\frac{2}{3} = ? - \frac{5}{3} + \frac{7}{3}

Add fractions

\frac{2}{3} = \frac{2}{3} ✓

By following the same procedure, the solution

x \approx 0

can be verified.

Solution	Substitute	Simplify
$x = - 1$	$2 (3)^{- 1} = ? \frac{5}{3} (- 1) + \frac{7}{3}$	$\frac{2}{3} = \frac{2}{3} ✓$
$x \approx 0$	$2 (3)^{0} \approx ? \frac{5}{3} (0) + \frac{7}{3}$	$2 \approx \frac{7}{3} ⇕ 2 \approx 2.333333 \dots ✓$

Since true statements were obtained, $x = - 1$ and $x \approx 0$ are solutions to the equation.

Example

Soccer or Football?

Dominika has physical education right before lunch. She and her teacher are both football and soccer fans. They know that, starting from

2020,

the attendance to the Major League Soccer's final game and the Super Bowl can be modeled by exponential functions.

MLS Final Game y = 40000 (1.25)^{x} Super Bowl y = 120000 (0.87)^{x}

In both cases,

x

is the number of years that have passed since

2020 .

Stadium

Dominika and her teacher want to find the year in which the attendance will be the same for both events. To do so, they need to solve an exponential equation by graphing. Help them find the year in which the attendance will be the same!

Hint

Graph $y = 40000 (1.25)^{x}$ and $y = 120000 (0.87)^{x}$ on the same coordinate plane. What is the $x -$ coordinate of the point of intersection? What does it mean in this context?

Solution

To find in which year the attendance to both events will be the same, Dominika needs to solve the exponential equation formed by the right-hand sides of the given functions.

40000 (1.25)^{x} = 120000 (0.87)^{x}

To solve the equation, two exponential functions will be considered.

40000 (1.25)^{x} = 120000 (0.87)^{x} ⇓ y = 40000 (1.25)^{x} y = 120000 (0.87)^{x} (I) (II)

These two functions will be graphed on the same coordinate plane. Their initial values and constant multipliers will be used to graph the functions.

The graphs intersect at one point. Although it can be seen that the $x -$ coordinate of the point of intersection is a bit greater than $3,$ its exact value cannot be determined by graphing.

Since $x$ is the number of years that have passed since the year $2020,$ it can be stated that the attendance to both events will be roughly the same in $2020 + 3 = 2023 .$

Discussion

Solving Exponential Equations Algebraically

Before discussing how to solve exponential equations algebraically, an important property must be learned.

Rule

Property of Equality for Exponential Equations

Two powers with the same positive base $b,$ where $b \neq = 1,$ are equal if and only if their exponents are equal.

If $b > 0$ and $b \neq = 1,$ then $b^{x} = b^{y}$ if and only if $x = y .$

Proof

This property will be proven in two parts.

If $b^{x} = b^{y},$ Then $x = y$

It is known that

b

is a positive number other than

1 .

This implies, among other things, that

b

is not zero. Consequently,

b^{y}

is never equal to zero and both sides of the equation can therefore be divided by

b^{y} .

Then, the Quotient of Powers Property can be used.

b^{x} = b^{y}

$LHS / b^{y} = RHS / b^{y}$

\frac{b ^{x}}{b ^{y}} = 1

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

b^{x - y} = 1

If a power with base

b \neq = 1

is equal to one, then the exponent is zero.

b^{x - y} = 1 \Rightarrow x - y = 0

The equation obtained means that

x

and

y

are equal.

x - y = 0 \Leftrightarrow x = y

It has been shown that if

b^{x} = b^{y},

then

x = y .

Note that if

b = 1,

the first implication is not valid because

1

raised to any power equals

1 .

In such a case,

x - y

would not be necessarily

0 .

This is why

b

must be a number other than

1!

If $x = y,$ Then $b^{x} = b^{y}$

Suppose for a moment that

b = 0 .

Now, raise

b

to a negative exponent

- n,

where

n

is a natural number.

b^{- n} \to 0^{- n}

Next, simplify the negative exponent and recall that

0^{n} = 0

for any natural number

n .

0^{- n} = \frac{1}{0 ^{n}} = \frac{1}{0}

Since division by zero is not defined, the expression

\frac{1}{0}

is not defined. This means that if

b = 0,

then it cannot be raised to a negative exponent. However, since in this case

b > 0,

it can be raised to any exponent

x

and the expression

b^{x}

will always be well defined.

b^{x} is well defined for b > 0

Now, by the Symmetric Property of Equality, write

b = b .

Then, use the above information to raise both sides of this equation to the power of

x .

Finally, use the fact that

x

and

y

are equal.

b = b

$LHS^{x} = RHS^{x}$

b^{x} = b^{x}

$x = y$

b^{x} = b^{y}

It has been shown that if

x = y,

then

b^{x} = b^{y} .

Therefore, the biconditional statement has been proven.

If $b > 0$ and $b \neq = 1,$ then $b^{x} = b^{y}$ if and only if $x = y .$

With this property in mind, a method for solving exponential equations algebraically can be explained.

Method

Solving Exponential Equations Algebraically

Let $b$ be a positive number other than $1$ and $a (x)$ and $c (x)$ be two algebraic expressions in terms of the same variable. If an exponential equation is or can be written in the following form, then it can be solved algebraically by using the Property of Equality for Exponential Equations.

$b^{a (x)} = b^{c (x)}$

Consider an example exponential equation.

4^{2 x} = 4096

To solve the equation, four steps must be followed.

1

Rewrite the Expressions on Both Sides as Powers With the Same Base

First, the exponential expressions on both sides must be rewritten as powers with the same base. In this case,

4096

can be written as a power of

4 .

4^{2 x} = 4096 \Leftrightarrow 4^{2 x} = 4^{6}

If the exponential expressions on both sides of the equation already have the same base, this step can be skipped.

2

Use the Property of Equality for Exponential Equations

Next, the Property of Equality for Exponential Equations can be used. Since the bases are equal, the exponents must be equal for the equation to be true.

4^{2 x} = 4^{6} \Leftrightarrow 2 x = 6

3

Solve the Resulting Equation

The resulting equation can be solved for the variable.

2 x = 6 \Leftrightarrow x = 3

4

Verify the Solution

Finally, the obtained solution can be verified by substituting it into the given equation.

4^{2 x} = 4096

$x = 3$

4^{2 (3)} = ? 4096

▼

Evaluate left-hand side

Multiply

4^{6} = ? 4096

Calculate power

4096 = 4096 ✓

Since a true statement was obtained,

x = 3

is a solution to the equation. It is important to verify all the obtained solutions, since sometimes this method can lead to extraneous solutions.

Example

Extra Credit Conundrums

Dominika decides to make good use of her free period after lunch to do some extra credit math problems.

smart

Unfortunately, she is struggling with solving three exponential equations. Help her understand how to solve the equations algebraically to obtain the extra credit she needs!

a

3^{3 x} = 3^{x + 1}

b

5^{x} = 2 5^{2 x + 1}

c

2 (2)^{x - 3} = (\frac{1}{8})^{\frac{1}{3} x}

Hint

a The exponential expressions on both sides of the equation already have the same base. Therefore, the Property of Equality for Exponential Equations can be used.

b Rewrite the expression on the right-hand side as a power with base

5 .

Then, use the Property of Equality for Exponential Equations.

c Rewrite the expressions on both sides as single powers with base

2 .

Then, use the Property of Equality for Exponential Equations.

Solution

a Since the exponential expressions on both sides of the equation already have the same base, the Property of Equality for Exponential Equations can be applied.

3^{3 x} = 3^{x + 1} \Leftrightarrow 3 x = x + 1

Next, the obtained equation can be solved for

x .

3 x = x + 1

▼

Solve for

x

$LHS - x = RHS - x$

2 x = 1

$LHS / 2 = RHS / 2$

x = \frac{1}{2}

The solution will now be verified by substituting

\frac{1}{2}

for

x

in the given equation.

3^{3 x} = 3^{x + 1}

$x = \frac{1}{2}$

3^{3 (\frac{1}{2})} = ? 3^{\frac{1}{2} + 1}

▼

Evaluate

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

3^{\frac{3}{2}} = ? 3^{\frac{1}{2} + 1}

Rewrite $1$ as $\frac{2}{2}$

3^{\frac{3}{2}} = ? 3^{\frac{1}{2} + \frac{2}{2}}

Add fractions

3^{\frac{3}{2}} = ? 3^{\frac{3}{2}}

Calculate power

5.196152 \dots = 5.196152 \dots ✓

Since a true statement was obtained,

x = \frac{1}{2}

is a solution to the equation.

b The right-hand side of this equation must be written as a power of

5 .

2 5^{2 x + 1}

▼

Rewrite

Write as a power

(5^{2})^{2 x + 1}

$(a^{m})^{n} = a^{m \cdot n}$

5^{2 (2 x + 1)}

Distribute $2$

5^{4 x + 2}

Now both sides of the equation are written as powers of

5 .

5^{x} = 2 5^{2 x + 1} \Leftrightarrow 5^{x} = 5^{4 x + 2}

Therefore, the Property of Equality for Exponential Equations will be used.

5^{x} = 5^{4 x + 2} \Leftrightarrow x = 4 x + 2

The value of

x

can be found by solving the equation.

x = 4 x + 2

▼

Solve for

x

$LHS - 4 x = RHS - 4 x$

- 3 x = 2

$LHS / (- 3) = RHS / (- 3)$

x = \frac{2}{- 3}

MoveNegDenomToFrac

Put minus sign in front of fraction

x = - \frac{2}{3}

The solution can now be checked by substituting

- \frac{2}{3}

for

x

in the given equation.

5^{x} = 2 5^{2 x + 1}

$x = - \frac{2}{3}$

5^{- \frac{2}{3}} = ? 2 5^{2 (- \frac{2}{3}) + 1}

▼

Evaluate

$a (- b) = - a \cdot b$

5^{- \frac{2}{3}} = ? 2 5^{- 2 (\frac{2}{3}) + 1}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

5^{- \frac{2}{3}} = ? 2 5^{- \frac{4}{3} + 1}

Rewrite $1$ as $\frac{3}{3}$

5^{- \frac{2}{3}} = ? 2 5^{- \frac{4}{3} + \frac{3}{3}}

Add fractions

5^{- \frac{2}{3}} = ? 2 5^{- \frac{1}{3}}

Calculate power

0.341995 \dots = 0.341995 \dots ✓

A true statement was obtained. Therefore,

x = - \frac{2}{3}

is a solution to the equation.

c In this equation, the expressions on both sides will be written as powers of

2 .

To do so, some Properties of Exponents will be used. The left-hand side will be rewritten first.

2 (2)^{x - 3}

▼

Rewrite

$a \cdot a^{m} = a^{1 + m}$

2^{1 + x - 3}

Subtract term

2^{x - 2}

Next, the right-hand side of the equation will be rewritten as a power of

2 .

(\frac{1}{8})^{\frac{1}{3} x}

▼

Rewrite

Write as a power

(\frac{1}{2 ^{3}})^{\frac{1}{3} x}

FracToNegExponent

$\frac{1}{a ^{m}} = a^{- m}$

(2^{- 3})^{\frac{1}{3} x}

$(a^{m})^{n} = a^{m \cdot n}$

2^{- 3 (\frac{1}{3} x)}

DenomMultFracToNumber

$3 \cdot \frac{a}{3} = a$

2^{- 1 x}

$(- a) b = - a b$

2^{- x}

The expressions on both sides of the equation are now written as powers of the same base.

2 (2)^{x - 3} = (\frac{1}{8})^{\frac{1}{3} x} \Leftrightarrow 2^{x - 2} = 2^{- x}

Therefore, the Property of Equality for Exponential Equations can be used.

2^{x - 2} = 2^{- x} \Leftrightarrow x - 2 = - x

Finally, the obtained equation can be solved for

x .

x - 2 = - x

▼

Solve for

x

$LHS + x = RHS + x$

2 x - 2 = 0

$LHS + 2 = RHS + 2$

2 x = 2

$LHS / 2 = RHS / 2$

x = 1

Now,

x = 1

will be checked by substitution.

2 (2)^{x - 3} = (\frac{1}{8})^{\frac{1}{3} x}

$x = 1$

2 (2)^{1 - 3} = ? (\frac{1}{8})^{\frac{1}{3} (1)}

▼

Evaluate

Subtract term

2 (2)^{- 2} = ? (\frac{1}{8})^{\frac{1}{3} (1)}

Identity Property of Multiplication

2 (2)^{- 2} = ? (\frac{1}{8})^{\frac{1}{3}}

Calculate power

2 (\frac{1}{4}) = ? \frac{1}{2}

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

\frac{2}{4} = ? \frac{1}{2}

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{2} = \frac{1}{2} ✓

A true statement was obtained. Therefore,

x = 1

is a solution to the equation.

Pop Quiz

Solving Exponential Equations

Solve the exponential equations graphically or algebraically. Whenever necessary, round the answers to two decimal places.

Exponential equations

Example

Estimating Bacterial Growth

Dominika finishes her Mondays with biology class. She learns that bacteria have the ability to multiply at incredible rates. After studying two bacteria populations, she concludes that their growth can be modeled by two exponential functions.

bacteria

In these formulas, the variable

y

is the number of bacteria in thousands

x

hours after the first observation. Dominika wants to know for how many hours after the first observation that population (I) will be less than population (II). To do so, she will solve the following exponential inequality.

10 (1.96)^{x} < 14 (1.4)^{x}

Help Dominika solve the inequality!

Hint

Rewrite the inequality so that each side is an exponential expression with the same base.

Solution

Exponential inequalities are solved the same way as exponential equations. To start, the inequality will be rewritten so that each side is an exponential expression with the same base.

10 (1.96)^{x} < 14 (1.4)^{x}

▼

Rewrite

$LHS / 10 < RHS / 10$

1.9 6^{x} < 1.4 (1.4)^{x}

Write as a power

(1 . 4^{2})^{x} < 1.4 (1.4)^{x}

$(a^{m})^{n} = a^{m \cdot n}$

1 . 4^{2 x} < 1.4 (1.4)^{x}

$a \cdot a^{m} = a^{1 + m}$

1 . 4^{2 x} < 1 . 4^{1 + x}

Now that both sides of the inequality are written as exponential expressions with the same base, the exponents can be compared.

1 . 4^{2 x} < 1 . 4^{1 + x} \Leftrightarrow 2 x < 1 + x

Finally, the obtained inequality can be solved for

x .

2 x < 1 + x \Leftrightarrow x < 1

It has been found that population (I) will be less than population (II) only for one hour. This can be verified by graphing each side of the inequality as an individual exponential function.

It is shown that the graph of $y = 10 (1.96)^{x}$ is below the graph of $y = 14 (1.4)^{x}$ for $x < 1,$ even before the first observation was made at $x = 0 .$ Therefore, the solution to the inequality is, indeed, $x < 1 .$

Closure

Finding the Year in Which the Cities Will Have the Same Population

With the topics seen in this lesson, the challenge presented at the beginning can be solved. Dominika has a meeting with her guidance counselor on Monday afternoon to discuss her college plans. Dominika wants to study in a smaller city. She knows that the populations of two cities in the US are modeled by two exponential functions.

Population of two cities described by the functions f(x)=8(1.0048)^x and g(x)=3.5(1.0067)^x

Here,

x

is the number of years that have passed since the year

2000 .

Furthermore,

f

and

g

are the populations in millions of people after

x

years. If they keep growing like this, in what year will the populations be the same? Round the answer to the nearest century.

Hint

Solve the exponential equation $8 (1.0048)^{x} = 3.5 (1.0067)^{x} .$

Solution

To find the year in which the populations will be the same, an exponential equation must be solved.

8 (1.0048)^{x} = 3.5 (1.0067)^{x}

Since the expressions on each side cannot be written as exponential expressions with the same base, the equation will be solved graphically. To do so, both functions will be graphed on the same coordinate plane by using their initial values and constant multipliers.

The $x -$ coordinate of the point of intersection cannot be determined precisely on the graph. However, it does show that, to the nearest hundred years, the $x -$ coordinate is $400 .$ This means that the cities will have the same population around the year $2400,$ many years after Dominika has finished college.

Since Dominika wants to study in a smaller city, she will talk to her counselor about the city with the smaller population during the period of time when she will be in college. Based on the graph, this means she will apply to a college in City B.

Solving Exponential Equations

Solving Exponential Equations

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