Learn on PengiBig Ideas Math, Algebra 2Chapter 6: Exponential and Logarithmic Functions

Lesson 6: Solving Exponential and Logarithmic Equations

In this Grade 8 lesson from Big Ideas Math Algebra 2, Chapter 6, students learn how to solve exponential and logarithmic equations using the Property of Equality for Exponential Equations, logarithms, and inverse properties. The lesson covers techniques such as rewriting equations with a common base, taking logarithms of both sides, and checking for extraneous solutions. Students also apply these skills to real-world problems like Newton's Law of Cooling to model and solve for unknown values in exponential contexts.

Section 1

Solving Exponential Equations

Property

An exponential equation is one in which the variable is part of an exponent. Many exponential equations can be solved by writing both sides of the equation as powers with the same base. In general, if two equivalent powers have the same base, then their exponents must be equal. If bm=bnb^m = b^n, then m=nm=n (for b>0,b1b > 0, b \neq 1).

Examples

  • To solve the equation 3x=813^x = 81, rewrite it as 3x=343^x = 3^4. Equating the exponents gives the solution x=4x=4.
  • For the equation 5x1=1255^{x-1} = 125, write it as 5x1=535^{x-1} = 5^3. Then, solve the simpler equation x1=3x-1 = 3 to get x=4x=4.

Section 2

Solving Exponential Equations Using Logarithms

Property

When it is not possible to write the expressions with the same base, take the common logarithm or natural logarithm of both sides once the exponential is isolated. Then use the Power Property to move the exponent to the front as a factor and solve for the variable.

Examples

  • Solve 3x=203^x = 20. Take the common log of both sides: log(3x)=log(20)\log(3^x) = \log(20). Using the Power Property, xlog3=log20x \log 3 = \log 20. So, x=log20log32.727x = \frac{\log 20}{\log 3} \approx 2.727.
  • Solve 4ex1=304e^{x-1} = 30. First, isolate the exponential: ex1=7.5e^{x-1} = 7.5. Take the natural log of both sides: ln(ex1)=ln(7.5)\ln(e^{x-1}) = \ln(7.5). This simplifies to x1=ln(7.5)x-1 = \ln(7.5), so x=ln(7.5)+13.015x = \ln(7.5) + 1 \approx 3.015.

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Chapter 6: Exponential and Logarithmic Functions

  1. Lesson 1

    Lesson 1: Exponential Growth and Decay Functions

  2. Lesson 2

    Lesson 2: The Natural Base e

  3. Lesson 3

    Lesson 3: Logarithms and Logarithmic Functions

  4. Lesson 4

    Lesson 4: Transformations of Exponential and Logarithmic Functions

  5. Lesson 5

    Lesson 5: Properties of Logarithms

  6. Lesson 6Current

    Lesson 6: Solving Exponential and Logarithmic Equations

  7. Lesson 7

    Lesson 7: Modeling with Exponential and Logarithmic Functions

Lesson overview

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Section 1

Solving Exponential Equations

Property

An exponential equation is one in which the variable is part of an exponent. Many exponential equations can be solved by writing both sides of the equation as powers with the same base. In general, if two equivalent powers have the same base, then their exponents must be equal. If bm=bnb^m = b^n, then m=nm=n (for b>0,b1b > 0, b \neq 1).

Examples

  • To solve the equation 3x=813^x = 81, rewrite it as 3x=343^x = 3^4. Equating the exponents gives the solution x=4x=4.
  • For the equation 5x1=1255^{x-1} = 125, write it as 5x1=535^{x-1} = 5^3. Then, solve the simpler equation x1=3x-1 = 3 to get x=4x=4.

Section 2

Solving Exponential Equations Using Logarithms

Property

When it is not possible to write the expressions with the same base, take the common logarithm or natural logarithm of both sides once the exponential is isolated. Then use the Power Property to move the exponent to the front as a factor and solve for the variable.

Examples

  • Solve 3x=203^x = 20. Take the common log of both sides: log(3x)=log(20)\log(3^x) = \log(20). Using the Power Property, xlog3=log20x \log 3 = \log 20. So, x=log20log32.727x = \frac{\log 20}{\log 3} \approx 2.727.
  • Solve 4ex1=304e^{x-1} = 30. First, isolate the exponential: ex1=7.5e^{x-1} = 7.5. Take the natural log of both sides: ln(ex1)=ln(7.5)\ln(e^{x-1}) = \ln(7.5). This simplifies to x1=ln(7.5)x-1 = \ln(7.5), so x=ln(7.5)+13.015x = \ln(7.5) + 1 \approx 3.015.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 6: Exponential and Logarithmic Functions

  1. Lesson 1

    Lesson 1: Exponential Growth and Decay Functions

  2. Lesson 2

    Lesson 2: The Natural Base e

  3. Lesson 3

    Lesson 3: Logarithms and Logarithmic Functions

  4. Lesson 4

    Lesson 4: Transformations of Exponential and Logarithmic Functions

  5. Lesson 5

    Lesson 5: Properties of Logarithms

  6. Lesson 6Current

    Lesson 6: Solving Exponential and Logarithmic Equations

  7. Lesson 7

    Lesson 7: Modeling with Exponential and Logarithmic Functions