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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.

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.

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.

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.

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}

Substitute

$x = 1$

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

▼

Evaluate

AddTerms

Add terms

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

CalcPow

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.

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 -

intercept is

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

CalcPow

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

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

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 .

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}

DivEqn

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

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

DivPow

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

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

RaiseEqn

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

b^{x} = b^{x}

Substitute

$x = y$

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.

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.

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

Solve the Resulting Equation

The resulting equation can be solved for the variable.

2 x = 6 \Leftrightarrow x = 3

Verify the Solution

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

4^{2 x} = 4096

Substitute

$x = 3$

4^{2 (3)} = ? 4096

▼

Evaluate left-hand side

Multiply

4^{6} = ? 4096

CalcPow

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.

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!

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

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

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

SubEqn

$LHS - x = RHS - x$

2 x = 1

DivEqn

$LHS / 2 = RHS / 2$

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}

Substitute

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

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}

OneToFrac

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

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

AddFrac

Add fractions

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

CalcPow

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

WritePow

Write as a power

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

PowPow

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

5^{2 (2 x + 1)}

Distr

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

SubEqn

$LHS - 4 x = RHS - 4 x$

- 3 x = 2

DivEqn

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

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}

Substitute

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

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

▼

Evaluate

MultPosNeg

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

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}

OneToFrac

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

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

AddFrac

Add fractions

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

CalcPow

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

MultBasePow

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

2^{1 + x - 3}

SubTerm

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

WritePow

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}

PowPow

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

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

DenomMultFracToNumber

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

2^{- 1 x}

MultNegPos

$(- a) b = - a b$

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

AddEqn

$LHS + x = RHS + x$

2 x - 2 = 0

AddEqn

$LHS + 2 = RHS + 2$

2 x = 2

DivEqn

{{ article.displayTitle }}

Catch-Up and Review

When Will the Cities Have the Same Population?

Exponential Equations

Solving Exponential Equations Graphically

Savings Account

Hint

Solution

More Than One Solution

Hint

Solution

Soccer or Football?

Hint

Solution

Solving Exponential Equations Algebraically

Property of Equality for Exponential Equations

Proof

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

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

Solving Exponential Equations Algebraically

Extra Credit Conundrums

Hint

Solution

{{ article.displayTitle }}

Catch-Up and Review

Hint

Solution

Hint

Solution

Hint

Solution

Proof

If bx=by, Then x=y

If x=y, Then bx=by

Hint

Solution

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

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