Lesson 14
Solving Exponential Equations
14.1: A Valid Solution? (5 minutes)
Warm-up
This warm-up gives students a chance to see and analyze a worked solution to an exponential equation before they solve some in the next activity. An exponential equation with base \(e\) was chosen purposefully to reinforce that \(e\) is just a number, and, as such, students can work with it the same way they have worked with equations of other bases previously.
Students should be familiar with the first and last steps in the worked solution, but the second step may give them pause because the exponent is a product (rather than just a variable or a number). Monitor for students who recognize that the product could be treated as a single object and who see the middle two lines as equivalent equations in exponential and logarithmic forms.
Launch
Give students a moment of quiet think time, and then ask them to share their thinking with a partner.
Student Facing
To solve the equation \(5 \boldcdot e^{3a} = 90\), Lin wrote the following:
\(\displaystyle \begin {align}5 \boldcdot e^{3a} &= 90\\ e^{3a} &= 18\\ 3a &= \log_e 18\\ a&=\frac {\log_e 18}{3}\\ \end {align}\)
Is her solution valid? Be prepared to explain what she did in each step to support your answer.
Student Response
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Activity Synthesis
Select students to explain Lin’s worked solution. If students struggle to explain the transition from \(e^{3a} = 18\) to \(3a = \log_e 18\), consider temporarily replacing the \(3a\) with a single variable, such as \(y\), and then ask students to interpret \(e^y = 18\) to \(y = \log_e 18\). Once students see these two equations as equivalent, given the definition of logarithm, then return to \(3a = \log_e 18\) and the division of each side by 3.
Some students may think that the equation is not completely solved because the logarithm is not evaluated. Remind students that a solution written in log notation is the exact solution, and that evaluating it would give us an approximation (a less-precise solution).
14.2: Natural Logarithm (10 minutes)
Activity
Building from the warm-up, in this activity students encounter more exponential equations with base \(e\). They learn that a logarithm with base \(e\) has a special name, the natural logarithm, and use the “ln” function on a calculator to estimate its value for different inputs.
To write equivalent equations in exponential form and logarithmic form, students look for and make use structures similar to those they previously encountered (MP7).
Launch
Display the table for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students brief quiet think time, and then invite students to share their observations.
Explain to students that the logarithm of an exponential expression with base \(e\) is called the natural logarithm, written as \(\ln\). Emphasize that \(\ln x\) means the same thing as \(\log_e x\).
Once students have completed the table, ask them to pause. Make sure students recognize that each pair of equations can be interpreted the same way we interpreted equations in other bases, and that the notation \(\ln \text{(some number)}\) can be thought of as the exponent to which we raise a base \(e\) to produce that number.
Give students access to a scientific calculator and ask them to locate the “e” and “ln” buttons on the calculator. Ask them to try evaluating \(\ln e\) and \(\ln 10\) and verify that they get 1 and approximately 2.302585.
Design Principle(s): Maximize meta-awareness
Supports accessibility for: Visual-spatial processing
Student Facing
- Complete the table with equivalent equations. The first row is completed for you.
exponential form logarithmic form a. \(e^0= 1\) \(\ln 1=0\) b. \(e^1= e\) c. \(e^\text{-1} = \frac{1}{e}\) d. \(\ln \frac{1}{e^2} = \text-2\) e. \(e^x = 10\) - Solve each equation by expressing the solution using \(\ln\) notation. Then, find the approximate value of the solution using the “ln” button on a calculator.
- \(e^m=20\)
- \(e^n=30\)
- \(e^p=7.5\)
Student Response
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Anticipated Misconceptions
Students may struggle getting started with the presence of the number \(e\). Ask them to look at the first row of the table. The reason \(\ln(1) = 0\) is that 0 is the exponent of \(e\) that gives 1. Ask, for \(\ln(e)\), what power of \(e\) gives a result of \(e\)? (1, since \(e^1=e\).)
Activity Synthesis
Invite students to share their solutions for the 3 unknown values. Resolve any discrepancies, such as any disagreements about approximate values determined using technology.
14.3: Solving Exponential Equations (20 minutes)
Activity
In this activity, students use logarithmic expressions to solve exponential equations. A variety of problems are chosen so that some can be solved through exponential reasoning, while others, in order to give an exact answer, require the use of logarithms.
Monitor for students who use the following strategies to solve the given equations:
- Reference the base 10 log table from an earlier lesson (for example: to find \(10^x = 10,\!000\), look to the last entry in the table, which gives the value for \(\log 10,\!000\)).
- Use exponential reasoning (for example: by thinking of 10,000 as \(10^4\) and reason that if \(10^{2x} = 10^4\), then \(2x = 4\) and \(x = 2\)).
- Use the definition of logarithm to rewrite equations (for example: if \(10^n = 315\), then \(n\) is the exponent to which 10 needs to be raised in order to give 315, so \(n = \log 315\)).
Launch
Supports accessibility for: Organization; Attention; Social-emotional skills
Student Facing
Without using a calculator, solve each equation. It is expected that some solutions will be expressed using log notation. Be prepared to explain your reasoning.
- \(10^x = 10,\!000\)
- \(5 \boldcdot 10^x = 500\)
- \(10^{(x+3)} = 10,\!000\)
- \(10^{2x} = 10,\!000\)
- \(10^x = 315\)
- \(2 \boldcdot 10^x = 800\)
- \(10^{(1.2x)} = 4,\!000\)
- \(7\boldcdot 10^{(0.5x)} = 70\)
- \(2 \boldcdot e^x=16\)
- \(10 \boldcdot e^{3x}=250\)
Student Response
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Student Facing
Are you ready for more?
- Solve the equations \(10^{n} = 16\) and \(10^{n} = 2\). Express your answers as logarithms.
- What is the relationship between these two solutions? Explain how you know.
Student Response
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Anticipated Misconceptions
For students who struggle solving the equations explicitly or in logarithmic form, recommend trying to isolate a power of 10 on one side (with the variable). Sometimes the other side can be written as a power of 10 and sometimes it cannot. When it can, this can be used to solve for the variable. When it cannot suggest that they think about the meaning of the base 10 logarithm and what it tells them.
Activity Synthesis
Focus the discussion on the different strategies students used to solve the equations. Invite previously identified students to share in the order given in the Activity Narrative.
During the discussion, help students connect the last few questions to the discussions in earlier lessons about exact solutions versus approximations. Also highlight the fact that the solution to several of these problems (for example, \(10^n = 315\) or any of the problems involving \(e\)) can only be expressed exactly using logarithms because neither the log table nor exponential reasoning will produce an exact value. On the other hand, using logarithms to solve the first few problems is not necessary. Rules of exponents are well suited for solving these problems.
Design Principle(s): Support sense-making
Lesson Synthesis
Lesson Synthesis
Make sure students understand these key takeaways:
- Properties of operations apply when we solve exponential equations, just as when we solved other equations in the past and whether or not logarithms are necessary. Maintaining equality between the two sides of the equations is essential.
- When the unknown value is an exponent, sometimes we can determine its value by reasoning (for example, if \(2 \boldcdot 10^y = 0.2\), then \(10^y = 0.1\), so \(y\) must be -1) or by estimation. But other times, it is difficult to “undo” exponentiation mentally, so we use logarithms to express that unknown exponent.
- The natural logarithm is not fundamentally different than other logarithms, even though the notation commonly used is \(\ln\) (though \(\log_e\) is also sometimes used), and the base involved is expressed as a letter instead of a number. The relationship between the base, the exponent, and the value of the exponential expression is still the same when the base is \(e\) as when it is 2, 5, 10.
14.4: Cool-down - Solve Some Equations (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
So far we have solved exponential equations by
- finding whole number powers of the base (for example, the solution of \(10^x = 100,\!000\) is 5)
- estimation (for example, the solution of \(10^x = 300\) is between 2 and 3)
- using a logarithm and approximating its value on a calculator (for example, the solution of \(10^x = 300\) is \(\log 300 \approx 2.48\))
Sometimes solving exponential equations takes additional reasoning. Here are a couple of examples.
\(\displaystyle \begin {align} 5 \boldcdot 10^x &= 45\\ 5 \boldcdot 10^x &= 45\\10^x &= 9\\x &=\log 9\\ \end {align}\)
\(\displaystyle \begin {align} 10^{(0.2t)} &= 1,\!000\\ 10^{(0.2t)} &= 10^3\\ 0.2t &= 3\\ t &= \frac {3}{0.2}\\ t &=15\\ \end {align}\)
In the first example, the power of 10 is multiplied by 5, so to find the value of \(x\) that makes this equation true each side was divided by 5. From there, the equation was rewritten as a logarithm, giving an exact value for \(x\).
In the second example, the expressions on each side of the equation were rewritten as powers of 10: \(10^{(0.2t)}=10^3\). This means that the exponent \(0.2t\) on one side and the 3 on the other side must be equal, and leads to a simpler expression to solve where we don't need to use a logarithm.
How do we solve an exponential equation with base \(e\), such as \(e^x = 5\)? We can express the solution using the natural logarithm, the logarithm for base \(e\). Natural logarithm is written as \(\ln\), or sometimes as \(\log_e\). Just like the equation \(10^2 =100 \) can be rewritten, in logarithmic form, as \(\log_{10}100 = 2\), the equation \(e^0 = 1\) and be rewritten as \(\ln 1 = 0\). Similarly, \(e^{\text-2} = \frac{1}{e^2}\) can be rewritten as \(\ln \frac{1}{e^2} = \text{-}2\).
All this means that we can solve \(e^x = 5\) by rewriting the equation as \(x = \ln 5\). This says that \(x\) is the exponent to which base \(e\) is raised to equal 5.
To estimate the size of \(\ln 5\), remember that \(e\) is about 2.7. Because 5 is greater than \(e^1\), this means that \(\ln 5\) is greater than 1. \(e^2\) is about \((2.7)^2\) or 7.3. Because 5 is less than \(e^2\), this means that \(\ln 5\) is less than 2. This suggests that \(\ln 5\) is between 1 and 2. Using a calculator we can check that \(\ln 5 \approx 1.61\).