Introduction

Welcome! In this lesson, we’ll focus on solving exponential and logarithmic equations and inequalities. These problems are a natural next step after learning about exponential and logarithmic functions and their properties. Here, the variable appears inside the exponent or inside a logarithm, which makes the solving process different from linear or quadratic equations. To solve them, we use the properties of exponents and logarithms, as well as the fact that they are inverse functions of each other.

Essential Knowledge from the CED

According to the AP Precalculus Course and Exam Description (CED), these are the key ideas you’ll be expected to understand about exponential functions for the exam:

Exponential and logarithmic equations and inequalities can be solved using properties of exponents, properties of logarithms, and the inverse relationship between exponential and logarithmic functions. These tools can simplify and solve equations and inequalities involving exponents and logarithms.
When solving exponential and logarithmic equations using analytical or graphical methods, results should be examined for extraneous solutions that are precluded by mathematical or contextual limitations.
Logarithms can be used to rewrite exponential expressions in different ways that may reveal useful information. For example, $b^{x} = c^{(l o g_{c} (b)) (x)}$
The function $f (x) = a \cdot b^{x - h} + k$ is a combination of additive transformations of an exponential function. The inverse of $y = f (x)$ can be found by performing the operations in reverse to undo the mapping.
The function $f (x) = a \cdot lo g_{b} (x - h) + k$ is a combination of additive transformations of a logarithmic function. The inverse of $y = f (x)$ can be found by performing the operations in reverse to undo the mapping.

Extraneous Solutions

When solving exponential and logarithmic equations, it’s important to check for extraneous solutions. These are solutions that may appear during the solving process but do not actually satisfy the original equation. Extraneous solutions often arise when applying properties of logarithms or exponents, especially if steps like squaring both sides, taking logs, or using inverse operations introduce values that are not valid in the original context.

For logarithmic equations, any solution that results in taking the logarithm of a non-positive number is invalid, because logarithms are only defined for positive arguments. Similarly, some exponential equations may produce results that don’t make sense in the context of the problem. Always substitute potential solutions back into the original equation to confirm which ones are valid.

Recap of Key Properties

Exponents:

Product Rule: $a^{m} \cdot a^{n} = a^{m + n}$
Quotient Rule: $\frac{a ^{m}}{a ^{n}} = a^{m - n}$
Power Rule: $(a^{m})^{n} = a^{m \cdot n}$
Negative Exponent Rule: $a^{- n} = \frac{1}{a ^{n}}$

Logarithms:

Product Rule: $lo g_{b} (M \cdot N) = lo g_{b} (M) + lo g_{b} (N)$
Quotient Rule: $lo g_{b} (\frac{M}{N}) = lo g_{b} (M) - lo g_{b} (N)$
Power Rule: $lo g_{b} (M^{p}) = p \cdot lo g_{b} (M)$
Change-of-Base Formula: $lo g_{b} (M) = \frac{lo g ( M )}{lo g ( b )}$

These properties are essential tools for solving exponential and logarithmic equations and inequalities, and they also help us catch extraneous solutions when checking our answers.

Solving Exponential Equations

Exponential equations are equations where the variable appears in the exponent. To solve them, we rely on properties of exponents and the inverse relationship between exponential and logarithmic functions.

Methods to solve exponential equations:

Same-Base Method
1. If both sides of the equation can be written with the same base, we can set the exponents equal. This works because of the one-to-one property of exponential functions: if $a^{m} = a^{n}$ , then $m = n$ .
2. Example: Solve $2^{3 x} = 16$
  1. Rewrite $16$ as $2^{4}$ using exponent rules
  2. Equation becomes $2^{3 x} = 2^{4}$
  3. Apply the one-to-one property: $3 x = 4 ⟹ x = \frac{4}{3}$
Using Logarithms
1. When the bases cannot be made the same, we can use logarithms to “bring down” the exponent. This uses the inverse relationship between exponentials and logarithms along with the power rule of logarithms.
2. Example: Solve $3^{x} = 20$
  1. Take the logarithm of both sides: $lo g (3^{x}) = lo g (20)$
  2. Apply the logarithm power rule: $x \cdot lo g (3) = lo g (20)$
  3. Solve for $x ⟹ x = \frac{lo g ( 20 )}{lo g ( 3 )}$
Rewriting Exponentials with a Different Base
1. Sometimes, you cannot write both sides of an exponential equation using the same base, and taking logarithms directly might feel confusing or messy. In these cases, you can rewrite one side of the equation using a different base. This helps because once both sides are expressed as powers of the same base, you can solve using the one-to-one property.
2. The general formula: $b^{x} = c^{l o g_{c} (b) \cdot x}$
  1. Here, you are expressing $b^{x}$ as a power of $c$ .
  2. $lo g_{c} (b)$ is the exponent you need on $c$ to get $b$ .
  3. Once rewritten, if the other side of the equation is already a power of $c$ , you can set the exponents equal.

Example:

Solve $2^{x} = 9$

Step 1: Choose a base to rewrite

The number $9$ can be written as a power of $3$ : $9 = 3^{2}$
To match the base, rewrite $2^{x}$ as a power of $3$ using the formula: $2^{x} = 3^{(l o g_{3} (2) \cdot x)}$

Step 2: Rewrite the equation with the same base

Original equation: $2^{x} = 9$
Rewritten: $3^{(l o g_{3} (2) \cdot x)} = 3^{2}$

Step 3: Apply the one-to-one property

Since the bases are the same, the exponents must be equal:
$lo g_{3} (2) \cdot x = 2$

Step 4: Solve for $x$

Divide both sides by $lo g_{3} (2)$ : $x = \frac{2}{lo g _{3} ( 2 )}$

Step 5: Check domain

$2^{x}$ is defined for all real $x$ , so no extraneous solutions here.

Solving Logarithmic Equations

Logarithmic equations are equations where the variable appears inside a logarithm. To solve them, we rely on properties of logarithms and the inverse relationship between logarithms and exponentials. Because logarithms are only defined for positive arguments, we must always check our solutions to make sure they are valid.

Using Logarithm Properties to Simplify

Sometimes an equation has multiple logarithms. We can use product, quotient, or power rules to combine them into a single logarithm, which makes it easier to solve.

Example:

Solve $lo g (x) + lo g (x - 2) = 1$

Step 1: Apply the product rule

The product rule states that $lo g (A) + lo g (B) = lo g (A \cdot B)$
Combine the logs: $lo g (x \cdot (x - 2)) = 1$

Step 2: Rewrite in exponential form

The inverse relationship of logs and exponentials says $lo g_{b} (M) = N$ is the same as $b^{N} = M$
Here, the base is $10$ (common log), so $1 0^{1} = x \cdot (x - 2) ⟹ 10 = x \cdot (x - 2)$

Step 3: Solve the resulting quadratic equation

Expand: $x^{2} - 2 x = 10 ⟹ x^{2} - 2 x - 10 = 0$
Now solve using the quadratic formula:
- $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
- Here, $a = 1$ , $b = - 2$ , $c = - 10$
Substitute:
- $x = \frac{2 \pm (( - 2 ) ^{2} - 4 ( 1 ) ( - 10 )))}{2}$
- $x = \frac{2 \pm 44}{2}$
- $x = \frac{2 \pm 2 11}{2} ⟹ x = 1 \pm 11$

Step 4: Check for extraneous solutions

Logarithms are only defined for positive arguments.
$x$ must be $> 0$ and $x - 2$ must be $> 0 ⟹ x > 2$
$x = 1 - 11$ is negative → extraneous solution
Valid solution: $x = 1 + 11$

Takeaway: Combining logs can introduce solutions that look mathematically correct but are not valid in the context of the logarithm. Always check the domain.

Isolating a Single Logarithm

If the equation has only one logarithm, we can isolate it and convert it to exponential form.

Example:

Solve $lo g_{2} (x - 3) = 4$

Step 1: Use the inverse relationship

$lo g_{b} (M) = N$ is equivalent to $b^{N} = M$
$2^{4} = x - 3 ⟹ 16 = x - 3$

Step 2: Solve for $x$

$x = 16 + 3 ⟹ x = 19$

Step 3: Check the domain

Argument $x - 3$ must be positive → $x > 3$
$x = 19$ satisfies this → valid solution

Logarithms with Different Bases

Sometimes logs have different bases or appear on both sides. We can use the change-of-base formula or rewrite in terms of the same base.

Example:

Solve $lo g_{3} (x) = lo g_{9} (27)$

Step 1: Rewrite the right-hand side using base $3$

$lo g_{9} (27) = \frac{lo g _{3} ( 27 )}{lo g _{3} ( 9 )} ⟹ \frac{lo g _{3} ( 27 )}{lo g _{3} ( 3 ^{2} )} ⟹ \frac{lo g _{3} ( 27 )}{2 \cdot lo g _{3} ( 3 )}$
$lo g_{3} (27) = 3$ because $3^{3} = 27$ , $lo g_{3} (3) = 1 ⟹ \frac{lo g _{3} ( 27 )}{2} = \frac{3}{2}$

Step 2: Solve for $x$

$lo g_{3} (x) = \frac{3}{2} ⟹ x = 3^{3/2} ⟹ x = 27$

Step 3: Check the domain

$x > 0$ , $27 > 0$ → valid solution

Solving Exponential Inequalities

Exponential inequalities are inequalities where the variable appears in the exponent, such as $2^{x} > 8$ or $3^{x - 1} \leq 9$ . Solving them is similar to solving exponential equations, but you must be careful with how you compare and test the values to find the correct solution.

Same-Base Method

Example:

Solve $2^{3 x} - 16 > 0$

Step 1: Solve the corresponding equation for the critical point

$2^{3 x} - 16 = 0 ⟹ 2^{3 x} = 16$
Rewrite $16$ as $2^{4}$ → $2^{3 x} = 2^{4}$
Apply one-to-one property → $3 x = 4$
Solve for $x$ → $x = \frac{4}{3}$

Step 2: Test intervals around the critical point

Pick $x = 0$ (less than $\frac{4}{3}$ ): $2^{(3 * 0)} - 16 = 1 - 16 = - 15$ → not $> 0$
Pick $x = 2$ (greater than $\frac{4}{3}$ ): $2^{(3 * 2)} - 16 = 64 - 16 = 48$ → satisfies $> 0$

Step 3: State solution

$x > \frac{4}{3}$

Rewriting Exponentials with a Different Base

Example:

Solve $3^{x} - 20 \leq 0$

Step 1: Solve the corresponding equation for the critical point

$3^{x} - 20 = 0 ⟹ 3^{x} = 20$
Take logarithms: $lo g (3^{x}) = lo g (20)$
Apply power rule: $x \cdot lo g (3) = lo g (20)$
Solve for $x$ → $x = \frac{lo g ( 20 )}{lo g ( 3 )} \approx 2.73$

Step 2: Test intervals around the critical point

$x = 0$ (less than $2.73$ ): $3^{0} - 20 = - 19$ → satisfies $\leq 0$
$x = 3$ (greater than $2.73$ ): $3^{3} - 20 = 7$ → does not satisfy

Step 3: State solution

$x \leq 2.73$

Rewriting Exponentials with a Different Base

Example:

Solve $2^{x} - 9 < 0$

Step 1: Solve the corresponding equation for the critical point

Start with the equation: $2^{x} - 9 = 0 ⟹ 2^{x} = 9$
Since the base $2$ cannot be rewritten as a neat power of $9$ , we use logarithms:
Take log of both sides: $lo g (2^{x}) = lo g (9)$
Apply the logarithm power rule: $x \cdot lo g (2) = lo g (9)$
Solve for $x$ : $x = \frac{lo g ( 9 )}{lo g ( 2 )} \approx 3.17$

Step 2: Determine intervals around the critical point

Critical point: $x \approx 3.17$
Intervals: $(- \infty, 3.17)$ and $(3.17, \infty)$

Step 3: Test numbers in each interval

Pick $x = 0$ (in $(- \infty, 3.17)$ ): $2^{0} - 9 = 1 - 9 = - 8$ → satisfies $< 0$
Pick $x = 4$ (in $(3.17, \infty)$ ): $2^{4} - 9 = 16 - 9 = 7$ → does not satisfy $< 0$

Step 4: State solution

Inequality is satisfied only when $x < 3.17$ → $x < \frac{lo g ( 9 )}{lo g ( 2 )}$

Solving Logarithmic Inequalities

Logarithmic inequalities are inequalities that involve logarithmic expressions, such as $lo g (x - 2) > 1$ or $ln (3 x + 1) \leq 2$ . Solving them is similar to solving logarithmic equations, but you also need to test intervals and check the domain to see where the inequality holds true.

Single Logarithm

Example:

Solve $lo g_{2} (x - 1) \geq 3$

Step 1: Solve the corresponding equation for the critical point

Set the inequality equal to find where it changes: $lo g_{2} (x - 1) = 3$
Rewrite in exponential form: $x - 1 = 2^{3}$
$x - 1 = 8$
$x = 9$
So the critical point is $x = 9$ .

Step 2: Determine Domain

The argument of a logarithm must be positive: $x - 1 > 0 ⟹ x > 1$

Step 3: Divide the number line into intervals

Based on the domain and critical point: $(1, 9)$ and $(9, \infty)$

Step 4: Test values from each interval

Pick $x = 5$ (in $(1, 9)$ ): $lo g_{2} (5 - 1) = lo g_{2} (4) = 2 ⟹ 2 \geq 3$ ✗ not true
Pick $x = 10$ (in $(9, \infty)$ ): $lo g_{2} (10 - 1) = lo g_{2} (9) \approx 3.17$ → $3.17 \geq 3$ ✓ true

Step 5: State the solution

The inequality is true for $x \geq 9$ , while also satisfying the domain $x > 1$ .

Using Log Properties (Multiple Logs Combined)

Example:

Solve $lo g (x) + lo g (x - 2) \leq 1$

Step 1: Determine domain

$x > 0$ and $x - 2 > 0$ → $x > 2$

Step 2: Combine logs using product rule

$lo g (x) + lo g (x - 2) = lo g (x \cdot (x - 2))$ → inequality: $lo g (x (x - 2)) \leq 1$

Step 3: Convert to exponential form

Base $10$ assumed → $x (x - 2) \leq 10$

Step 4: Solve the corresponding equation for the critical point(s)

Set it equal first to find where the inequality changes: $x (x - 2) = 10$
Expand: $x^{2} - 2 x - 10 = 0$
Now solve using the quadratic formula:
- $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
- Here, $a = 1$ , $b = - 2$ , $c = - 10$
Substitute:
- $x = \frac{2 \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - 10 )}{2}$
- $x = \frac{2 \pm 4 + 40}{2}$
- $x = \frac{( 2 \pm 44 )}{2} = \frac{2 \pm 2 11}{2} ⟹ x = 1 \pm 11$

Step 5: Check for extraneous solutions

From Step 1, the domain is $x > 2$ , so $x \approx - 2.32$ is invalid. Only $x \approx 4.32$ is valid as a critical point.

Step 6: Divide number line into intervals and test

Now test intervals around the valid critical point
Test $x = 3$ (in $(2, 4.32)$ ): $3 (3 - 2) = 3 ⟹ 3 \leq 10$ ✓ satisfies
Test $x = 5$ (greater than $4.32$ ): $5 (5 - 2) = 15 ⟹ 15 \leq 10$ ✗ does not satisfy

Step 7: State the solution

The inequality is true for $x > 2$ and $x \leq 4.32$

Logarithms with Different Bases

Example:

Solve $lo g_{3} (x) > lo g_{9} (4)$

Step 1: Rewrite logs with the same base

$lo g_{9} (4) = \frac{lo g _{3} ( 4 )}{lo g _{3} ( 9 )} = \frac{lo g _{3} ( 4 )}{2}$

Step 2: Solve the corresponding equation for the critical point

$lo g_{3} (x) = \frac{lo g _{3} ( 4 )}{2} ⟹ x = 3^{l o g_{3} (4) /2} = 4 = 2$

Step 3: Determine domain

$x > 0$

Step 4: Divide number line into intervals

Intervals: $(0, 2)$ and $(2, \infty)$

Step 5: Test numbers in each interval

Pick x = 1 (in $(0, 2)$ ): $lo g_{3} (1) = 0 ⟹ 0 > lo g_{9} (4) \approx 0.63$ ✗ does not satisfy
Pick $x = 3$ (in $(2, \infty)$ ): $lo g_{3} (3) = 1 ⟹ 1 > 0.63$ ✓ satisfies

Step 6: State solution

$x > 2$

Inverses of Exponential and Logarithmic Functions

Both exponential and logarithmic functions can be transformed by shifting, stretching, or reflecting. Their inverses are found by undoing those same transformations in reverse order.

Exponential Functions:

General Form: $f (x) = a \cdot b^{x - h} + k$

$a$ → vertical stretch or reflection
$b$ → base of the exponential (growth if $b > 1$ , decay if $0 < b < 1$ )
$h$ → horizontal shift
$k$ → vertical shift

Example:

$f (x) = 2^{x - 1} + 3$

Find the inverse:

Replace $f (x)$ with $y$ → $y = 2^{x - 1} + 3$
Swap $x$ and $y$ → $x = 2^{y - 1} + 3$
Subtract $3$ → $x - 3 = 2^{y - 1}$
Take log base $2$ → $lo g_{2} (x - 3) = y - 1$
Add $1$ → $y = lo g_{2} (x - 3) + 1$

Inverse: $f^{- 1} (x) = lo g_{2} (x - 3) + 1$

Logarithmic Functions:

General Form: $f (x) = a \cdot lo g_{b} (x - h) + k$

$a$ → vertical stretch or reflection
$b$ → base of the logarithm
$h$ → horizontal shift
$k$ → vertical shift

Example:

$f (x) = lo g_{2} (x - 4) + 1$

Find the inverse:

Replace $f (x)$ with $y$ → $y = lo g_{2} (x - 4) + 1$
Swap $x$ and $y$ → $x = lo g_{2} (y - 4) + 1$
Subtract $1$ → $x - 1 = lo g_{2} (y - 4)$
Rewrite in exponential form → $2^{x - 1} = y - 4$
Add $4$ → $y = 2^{x - 1} + 4$

Inverse: $f^{- 1} (x) = 2^{x - 1} + 4$

FiveHive

FiveHive

2.13 - Exponential and Logarithmic Equations and Inequalities

Introduction

Essential Knowledge from the CED

Extraneous Solutions

Recap of Key Properties

Solving Exponential Equations

Solving Logarithmic Equations

Using Logarithm Properties to Simplify

Isolating a Single Logarithm

Logarithms with Different Bases

Solving Exponential Inequalities

Same-Base Method

Rewriting Exponentials with a Different Base

Rewriting Exponentials with a Different Base

Solving Logarithmic Inequalities

Single Logarithm

Using Log Properties (Multiple Logs Combined)

Logarithms with Different Bases

Inverses of Exponential and Logarithmic Functions

Exponential Functions:

Logarithmic Functions:

Practice Questions