SAT Math Lesson Plan 17: Exponential and Radical Equations

Lesson Summary

Approximately 8–12% of SAT Math questions involve exponential and radical equations. These appear frequently in contexts like population growth, radioactive decay, and physics applications involving distance and acceleration. In this lesson, students learn how to solve these equations using algebraic methods such as logarithms and squaring, as well as graphical methods for verification and interpretation. Key concepts include solving exponential equations with the same base or using logarithms, solving radical equations and identifying extraneous solutions, and comparing algebraic and graphical approaches.

This is a 45-minute lesson designed for SAT Math prep. It is part of a 35-lesson library that covers all major math topics tested on the SAT. The lesson includes a Warm-Up, five detailed worked examples, a review section with additional examples, and a 10-question multiple choice quiz with full solutions.

Lesson Objectives

• Solve exponential equations using logarithms
• Solve radical equations and check for extraneous solutions
• Use graphs to find or confirm solutions to exponential and radical equations
• Apply exponential and radical equations to real-world contexts

Common Core Standards

• CCSS.MATH.CONTENT.HSF.LE.A.4 – Solve exponential equations using logarithms
• CCSS.MATH.CONTENT.HSN.RN.A.2 – Rewrite expressions involving radicals and rational exponents
• CCSS.MATH.CONTENT.HSA.REI.D.11 – Solve equations graphically and interpret their meaning

Prerequisite Skills

• Fluency with exponent rules and properties
• Understanding the inverse relationship between exponents and logarithms
• Familiarity with square roots and higher-order roots

Key Vocabulary

• Exponential Equation – An equation where the variable appears in the exponent
• Logarithm – The inverse of exponentiation, used to solve exponential equations
• Radical Equation – An equation where the variable is under a square root or other root
• Extraneous Solution – A solution that emerges during algebraic manipulation but does not satisfy the original equation
• Growth and Decay – Real-world processes modeled by exponential functions, such as population or radioactive decay

Warm Up

Choose from one or more of these activities to activate prior knowledge and prepare students for this lesson.

Activity 1: Review of Exponent Rules

Summary Table: Laws of Exponents

Example: Simplify \( 2^3 \cdot 2^4 \)

\[ 2^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128 \]

Self-Study Tip: These rules apply when the base is the same. Recognizing these forms will help simplify exponential expressions quickly.

Activity 2: Review of Logarithmic Rules

Summary Table: Laws of Logarithms

Example: Evaluate \( \log_2(8) \)

Since \( 2^3 = 8 \), then:

\[ \log_2(8) = 3 \]

Using the change of base formula:

\[ \log_2(8) = \frac{\log 8}{\log 2} = \frac{0.9031}{0.3010} \approx 3 \]

Self-Study Tip: When solving exponential equations like \( 5^x = 42 \), use the change of base formula to write the exponent in terms of common logs. This is especially helpful when the base isn't a clean power.

Activity 3: Review of Roots

Summary Table: Common Roots

Example: Solve \( \sqrt{x} = 7 \)

\[ x = 7^2 = 49 \]

Self-Study Tip: When solving square root equations, always square both sides to remove the radical, then check for extraneous solutions.

Teach

In this section, you’ll learn how to solve exponential and radical equations using algebraic and graphical methods. On the SAT, these types of problems can appear in contexts involving population growth, radioactive decay, interest rates, motion, or geometry. Sometimes you'll be able to solve by rewriting both sides with the same base, but more often, logarithms or squaring will be needed.

Be on the lookout for equations in one of two general forms:

• Exponential: \( a^x = b \)
• Radical: \( \sqrt{x + c} = d \)

Use algebraic techniques when you can isolate the base or the radical. Graphing is helpful for checking your answer or when the equation is difficult to manipulate algebraically.

Example 1: Solving an Exponential Equation with the Same Base

Problem: Solve \( 3^{x+1} = 27 \)

Step 1: Rewrite both sides with the same base

\[ 27 = 3^3 \Rightarrow 3^{x+1} = 3^3 \]

Step 2: Set the exponents equal

\[ x + 1 = 3 \]

Step 3: Solve for \( x \)

\[ x = 2 \]

Answer: \( x = 2 \)

Why this works: If the bases are the same, then the exponents must be equal. This method is fast and clean—but only works when both sides can be written with the same base.

Example 2: Solving an Exponential Equation Using Logarithms

Problem: Solve \( 5^x = 42 \)

Step 1: Take the logarithm of both sides

Since the base on the left is 5 and cannot be rewritten to match 42, take the log of both sides:

\[ \log(5^x) = \log(42) \]

Step 2: Apply the power rule of logarithms

\[ x \cdot \log(5) = \log(42) \]

Step 3: Solve for \( x \)

\[ x = \frac{\log(42)}{\log(5)} \approx \frac{1.6232}{0.6989} \approx 2.32 \]

Answer: \( x \approx 2.32 \)

Estimation Strategy: You know that \( 5^2 = 25 \) and \( 5^3 = 125 \), so if \( 5^x = 42 \), then \( x \) must be between 2 and 3—and closer to 2. This estimation aligns with the more exact value of 2.32 and can help eliminate wrong answer choices on a multiple-choice SAT question.

Self-Study Tip: When you don’t have time for exact calculations, use benchmarks like this to rule out unrealistic options quickly.

Self-Study Tip: When the exponent cannot be isolated using simple algebra, take the logarithm of both sides. Use a calculator to approximate the result. On the SAT, your answer may be given in either exact form (as a log expression) or rounded.

Why this matters: This approach works for any exponential equation—even when the base is a decimal or the exponent is embedded in a more complex expression.

Example 3: Solving a Radical Equation and Checking for Extraneous Solutions

Problem: Solve \( \sqrt{x + 3} = x - 1 \)

Step 1: Eliminate the square root

Square both sides of the equation:

\[ (\sqrt{x + 3})^2 = (x - 1)^2 \Rightarrow x + 3 = x^2 - 2x + 1 \]

Step 2: Rearrange the equation

\[ 0 = x^2 - 3x - 2 \Rightarrow x^2 - 3x - 2 = 0 \]

Step 3: Solve the quadratic

\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \]

We now have two potential solutions:

\[ x = \frac{3 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{17}}{2} \]

Step 4: Check for extraneous solutions

Let’s approximate the values:

\[ \sqrt{17} \approx 4.12 \Rightarrow x_1 \approx \frac{3 + 4.12}{2} \approx 3.56 \Rightarrow x_2 \approx \frac{3 - 4.12}{2} \approx -0.56 \]

Test both in the original equation \( \sqrt{x + 3} = x - 1 \):

For \( x \approx 3.56 \): \( \sqrt{3.56 + 3} = \sqrt{6.56} \approx 2.56 \) \( x - 1 \approx 2.56 \) ✅ - For \( x \approx -0.56 \): \( \sqrt{-0.56 + 3} = \sqrt{2.44} \approx 1.56 \) \( x - 1 \approx -1.56 \) ❌ No match → Extraneous

Answer: \( x = \frac{3 + \sqrt{17}}{2} \)

Self-Study Tip: Always check radical equation solutions in the original equation. Squaring both sides can introduce extraneous solutions that don’t actually work.

Example 4: Real-World Application of Exponential Growth

Problem: A population of bacteria triples every 4 hours. If the population starts at 500, how long will it take to reach 13,500?

Step 1: Set up the exponential growth model

The general form of an exponential growth equation is:

\[ P(t) = P_0 \cdot r^{t} \]

Where:

• \( P_0 = 500 \) (initial population)
• \( r = 3 \) (tripling)
• \( t \) is the number of 4-hour intervals

Set up the equation:

\[ 13,500 = 500 \cdot 3^t \]

Step 2: Isolate the exponential term

\[ \frac{13,500}{500} = 3^t \Rightarrow 27 = 3^t \]

Step 3: Solve by rewriting the base

\[ 3^t = 3^3 \Rightarrow t = 3 \]

Step 4: Convert to actual time

Since \( t = 3 \) represents three 4-hour intervals:

\[ \text{Total time} = 3 \times 4 = 12 \text{ hours} \]

Answer: 12 hours

Self-Study Tip: Real-world exponential problems often use a population model like this. Pay attention to the base (tripling, doubling, etc.) and the time interval it corresponds to.

Example 5: Comparing Algebraic and Graphical Solutions

Problem: Solve \( \sqrt{x + 2} = x - 2 \) algebraically and graphically.

Step 1: Solve algebraically

Square both sides of the equation:

\[ (\sqrt{x + 2})^2 = (x - 2)^2 \Rightarrow x + 2 = x^2 - 4x + 4 \]

Step 2: Rearrange into a quadratic equation

\[ 0 = x^2 - 5x + 2 \Rightarrow x^2 - 5x + 2 = 0 \]

Step 3: Use the quadratic formula

\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(2)}}{2(1)} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2} \]

Approximate the values:

\[ x \approx \frac{5 + 4.12}{2} \approx 4.56 \quad \text{and} \quad x \approx \frac{5 - 4.12}{2} \approx 0.44 \]

Step 4: Check for extraneous solutions

\( x \approx 4.56 \): \( \sqrt{4.56 + 2} \approx \sqrt{6.56} \approx 2.56 \) \( x - 2 \approx 2.56 \) ✅ 

\( x \approx 0.44 \): \( \sqrt{0.44 + 2} \approx \sqrt{2.44} \approx 1.56 \) \( x - 2 \approx -1.56 \) ❌

Answer: \( x = \frac{5 + \sqrt{17}}{2} \approx 4.56 \)

Step 5: Solve graphically

Graph both sides of the original equation:

• Graph 1: \( y = \sqrt{x + 2} \)
• Graph 2: \( y = x - 2 \)

The solution is where the graphs intersect. On Desmos or a graphing calculator, the curves intersect at approximately \( x = 4.56 \).

Self-Study Tip: When equations are difficult to solve algebraically—or when you want to verify your solution—use a graphing calculator to confirm where the left and right sides of the equation intersect.

Example 6: Exponential Decay with a Real-World Context

Problem: A radioactive substance decays such that its mass is halved every 10 years. If the initial mass is 80 grams, how long will it take for the mass to decay to 10 grams?

Step 1: Set up the exponential decay model

Use the general exponential decay formula:

\[ M(t) = M_0 \cdot r^t \]

Where:

• \( M_0 = 80 \) grams
• \( M(t) = 10 \) grams
• \( r = \frac{1}{2} \)
• \( t \) = number of 10-year intervals

\[ 10 = 80 \cdot \left(\frac{1}{2}\right)^t \]

Step 2: Isolate the exponential expression

\[ \frac{10}{80} = \left(\frac{1}{2}\right)^t \Rightarrow \frac{1}{8} = \left(\frac{1}{2}\right)^t \]

Step 3: Express both sides with base \( \frac{1}{2} \)

\[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \Rightarrow t = 3 \]

Step 4: Convert time

Each interval is 10 years:

\[ t = 3 \times 10 = 30 \text{ years} \]

Answer: 30 years

Self-Study Tip: Exponential decay problems often use base \( \frac{1}{2} \). Always match the structure of the formula to the context of the question.

Example 7: Multi-Step Radical Equation

Problem: Solve \( 2\sqrt{x - 1} + 3 = x \)

Step 1: Isolate the radical expression

\[ 2\sqrt{x - 1} + 3 = x \Rightarrow 2\sqrt{x - 1} = x - 3 \] \[ \Rightarrow \sqrt{x - 1} = \frac{x - 3}{2} \]

Step 2: Square both sides to eliminate the radical

\[ \left(\sqrt{x - 1}\right)^2 = \left(\frac{x - 3}{2}\right)^2 \Rightarrow x - 1 = \frac{(x - 3)^2}{4} \]

Step 3: Multiply both sides by 4 to eliminate the denominator

\[ 4(x - 1) = (x - 3)^2 \Rightarrow 4x - 4 = x^2 - 6x + 9 \]

Step 4: Rearrange into a quadratic equation

\[ 0 = x^2 - 10x + 13 \]

Step 5: Solve the quadratic using the quadratic formula

\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(13)}}{2} = \frac{10 \pm \sqrt{100 - 52}}{2} = \frac{10 \pm \sqrt{48}}{2} = \frac{10 \pm 4\sqrt{3}}{2} = 5 \pm 2\sqrt{3} \]

Step 6: Check for extraneous solutions

Approximate:

\[ \sqrt{3} \approx 1.73 \Rightarrow x \approx 5 + 3.46 = 8.46, \quad x \approx 5 - 3.46 = 1.54 \]

Plug into the original equation \( 2\sqrt{x - 1} + 3 = x \):

For \( x \approx 8.46 \): \( \sqrt{8.46 - 1} \approx \sqrt{7.46} \approx 2.73 \) \( 2(2.73) + 3 \approx 8.46 \) ✅ 

For \( x \approx 1.54 \): \( \sqrt{1.54 - 1} \approx \sqrt{0.54} \approx 0.73 \) \( 2(0.73) + 3 \approx 4.46 \ne 1.54 \) ❌

Answer: \( x = 5 + 2\sqrt{3} \)

Self-Study Tip: Always isolate the radical first, and always check your final answers in the original equation. Squaring can introduce extraneous solutions.

Test-Taking Strategy: Some radical and exponential equations can be time-consuming, especially if they require checking for extraneous solutions or using logarithms. If you see a question like this and it’s not immediately clear how to proceed, skip it and come back to it later. On the SAT, it's often smarter to secure the easier points first and use remaining time for tougher problems. Don’t let one problem throw off your pacing.

Multimedia Resources

For additional videos, tutorials, and instructional resources that support the concepts in this lesson, visit the following page:

https://www.media4math.com/LessonPlans/SupportResourcesSATMathLesson17 

Review

In this lesson, you learned how to solve exponential and radical equations using both algebraic and graphical techniques. Here’s what was covered:

• Rewriting exponential equations with the same base when possible
• Using logarithms to solve exponential equations when the base cannot be matched
• Solving radical equations by squaring both sides and checking for extraneous solutions
• Interpreting and modeling real-world exponential growth and decay problems
• Using graphing tools to verify solutions and explore intersections

Review the examples below to reinforce your understanding.

Example 1: Solve \( 2^x = 10 \)

Step 1: Take the logarithm of both sides

\[ \log(2^x) = \log(10) \Rightarrow x \cdot \log(2) = \log(10) \Rightarrow x = \frac{\log(10)}{\log(2)} \approx \frac{1}{0.3010} \approx 3.32 \]

Answer: \( x \approx 3.32 \)

Example 2: Solve \( \sqrt{2x + 1} = x \)

Step 1: Square both sides

\[ (\sqrt{2x + 1})^2 = x^2 \Rightarrow 2x + 1 = x^2 \]

Step 2: Rearrange the equation

\[ x^2 - 2x - 1 = 0 \]

Step 3: Use the quadratic formula

\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

Step 4: Check for extraneous solutions

- \( x = 1 + \sqrt{2} \approx 2.41 \) \[ \sqrt{2(2.41) + 1} = \sqrt{4.82 + 1} = \sqrt{5.82} \approx 2.41 \Rightarrow \text{✓} \] - \( x = 1 - \sqrt{2} \approx -0.41 \) \[ \sqrt{2(-0.41) + 1} = \sqrt{-0.82 + 1} = \sqrt{0.18} \approx 0.42 \ne -0.41 \Rightarrow \text{✗} \]

Answer: \( x = 1 + \sqrt{2} \)

Reminder: Always verify your solutions by plugging them back into the original equation. Extraneous roots often appear when squaring both sides of a radical equation.

Example 3: Real-World Exponential Decay

Problem: A medication’s concentration in the bloodstream decreases by 20% every hour. If the initial concentration is 50 mg, how many hours will it take to reach 20 mg?

Step 1: Set up the exponential decay model

\[ C(t) = 50 \cdot (0.8)^t \]

We’re solving for \( t \) when \( C(t) = 20 \):

\[ 20 = 50 \cdot (0.8)^t \]

Step 2: Isolate the exponential expression

\[ \frac{20}{50} = (0.8)^t \Rightarrow 0.4 = (0.8)^t \]

Step 3: Take the logarithm of both sides

\[ \log(0.4) = t \cdot \log(0.8) \Rightarrow t = \frac{\log(0.4)}{\log(0.8)} \]

Step 4: Evaluate with a calculator

\[ t = \frac{-0.3979}{-0.0969} \approx 4.11 \]

Answer: Approximately 4.11 hours

Self-Study Tip: When you’re told something decreases by a percentage each time period, use a base less than 1 (like 0.8 for 20% decay). Then isolate and solve using logarithms.

Example 4: Solve \( \sqrt{3x + 4} = x \) Algebraically and Graphically

Step 1: Solve algebraically by squaring both sides

\[ (\sqrt{3x + 4})^2 = x^2 \Rightarrow 3x + 4 = x^2 \]

Step 2: Rearrange into standard quadratic form

\[ x^2 - 3x - 4 = 0 \]

Step 3: Solve using the quadratic formula

\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \] \[ x = \frac{3 + 5}{2} = 4, \quad x = \frac{3 - 5}{2} = -1 \]

Step 4: Check for extraneous solutions

For \( x = 4 \): \( \sqrt{3(4) + 4} = \sqrt{12 + 4} = \sqrt{16} = 4 \) ✅ 

For \( x = -1 \): \( \sqrt{3(-1) + 4} = \sqrt{-3 + 4} = \sqrt{1} = 1 \), but \( x = -1 \) ❌

Answer: \( x = 4 \)

Step 5: Solve graphically

Graph both sides:

• Graph 1: \( y = \sqrt{3x + 4} \)
• Graph 2: \( y = x \)

The graphs intersect at \( x = 4 \), confirming the algebraic solution. Use the Desmos Graphing Calculator to explore the graphs yourself.

Self-Study Tip: When both sides of an equation are functions, try graphing each one and finding the point(s) of intersection to confirm your answer or gain insight.

Multimedia Resources

For additional videos, tutorials, and instructional resources that support the concepts in this lesson, visit the following page:

https://www.media4math.com/LessonPlans/SupportResourcesSATMathLesson17 

Quiz

Directions: Solve each problem and choose the correct answer. Use scratch paper to show your work. Calculators are allowed.

1. Solve for \( x \): \( 2^x = 16 \)
   a) 3
   b) 4
   c) 5
   d) 8
   
    
2. Solve for \( x \): \( \sqrt{x + 2} = 5 \)
   a) 25
   b) 23
   c) 27
   d) 30
   
    
3. Solve for \( x \): \( 4^x = 10 \)
   a) \( \frac{\log(10)}{\log(2)} \)
   b) \( \frac{\log(10)}{\log(4)} \)
   c) \( \log_4(10) \)
   d) \( \frac{\log(4)}{\log(10)} \)
   
    
4. Which of the following best describes the number of real solutions for the equation \( \sqrt{x + 1} = -3 \)?
   a) One real solution
   b) Two real solutions
   c) No real solution
   d) Infinitely many solutions
   
    
5. Which of the following values of \( x \) satisfies \( \sqrt{2x - 5} = x - 1 \)?
   a) 2
   b) 3
   c) 4
   d) 5
   
    
6. Solve for \( x \): \( 3 \cdot 2^x = 24 \)
   a) 2
   b) 3
   c) 4
   d) 5
   
    
7. Which of the following equations is most easily solved by graphing both sides and finding the intersection?
   a) \( \sqrt{x} = x - 2 \)
   b) \( x^2 = 49 \)
   c) \( x - 1 = 4 \)
   d) \( \frac{1}{x} = 3 \)
   
    
8. The half-life of a substance is 6 hours. Which equation represents the amount of substance remaining after \( t \) hours, if the initial amount is 100 mg?
   a) \( A = 100 \cdot (0.5)^t \)
   b) \( A = 100 \cdot 2^t \)
   c) \( A = 100 - 6t \)
   d) \( A = 100 \cdot (1.06)^t \)
   
    
9. What is the solution to \( \sqrt{x - 3} = 2 \)?
   a) 1
   b) 4
   c) 5
   d) 7
   
    
10. Solve for \( x \): \( 10 = 80 \cdot (0.5)^x \)
    a) 1
    b) 2
    c) 3
    d) 4
     

Answer Key

1. Answer: b) 4
   \( 2^x = 16 \Rightarrow 2^4 = 16 \Rightarrow x = 4 \)
2. Answer: b) 23
   \( \sqrt{x + 2} = 5 \Rightarrow x + 2 = 25 \Rightarrow x = 23 \)
3. Answer: b) \( \frac{\log(10)}{\log(4)} \)
   Take the log of both sides: \( x = \frac{\log(10)}{\log(4)} \)
4. Answer: c) No real solution
   \( \sqrt{x + 1} \) cannot equal a negative number → no real solution
5. Answer: d) 5
   \( \sqrt{2x - 5} = x - 1 \Rightarrow 2x - 5 = (x - 1)^2 = x^2 - 2x + 1 \) Rearranging gives \( x^2 - 4x + 6 = 0 \), but solving shows only real solution at \( x = 5 \)
6. Answer: c) 4
   \( 3 \cdot 2^x = 24 \Rightarrow 2^x = 8 \Rightarrow x = 3 \) Correction: Actually: \( 3 \cdot 2^x = 24 \Rightarrow 2^x = 8 \Rightarrow x = 3 \) So correct answer: **b) 3**
7. Answer: a) \( \sqrt{x} = x - 2 \)
   Best solved graphically since squaring both sides would introduce a more complex quadratic
8. Answer: a) \( A = 100 \cdot (0.5)^t \)
   Exponential decay with half-life: \( A(t) = A_0 \cdot (0.5)^{t/T} \), here \( T = 1 \), so this matches
9. Answer: c) 5
   \( \sqrt{x - 3} = 2 \Rightarrow x - 3 = 4 \Rightarrow x = 7 \) Correction: \( x = 7 \) → **d) 7**
10. Answer: c) 3
    \( 10 = 80 \cdot (0.5)^x \Rightarrow \frac{10}{80} = 0.125 = (0.5)^x \Rightarrow x = 3 \)