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Precalculus for Team-Based Inquiry Learning 2025 Edition — Instructor Version

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Section 5.5 Properties of Logarithms (EL5)

Subsection 5.5.1 Activities

Remark 5.5.1.

Recall from Remark 5.3.5 that we can convert between exponential and logarithmic forms.
\begin{equation*} \log_bx=y \end{equation*}
is equivalent to
\begin{equation*} b^y=x\text{.} \end{equation*}
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Activity 5.5.2.

Suppose you are given two equations \(\log_bM=x\) and \(\log_bN=x\text{.}\)
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(a)
Rewrite each logarithmic equation into an exponential equation.
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Answer.
\(\log_bM=x \iff b^{x}=M \)
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\(\log_bN=x \iff b^{x}=N \)
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(b)
Look at the exponential equations you found in part (a). What conclusion can you make about \(M\) and \(N\text{?}\)
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Answer.
Because \(b^{x}=M\) and \(b^{x}=N\text{,}\) then it follows that \(M=N\text{.}\)
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(c)
Given that both \(\log_bM\) and \(\log_bN\) are both equal to \(x\text{,}\) what can you say about \(\log_bM\) and \(\log_bN\text{?}\)
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Answer.
\(\log_bM=\log_bN\)
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(d)
If given \(\log_bM=\log_bN\text{,}\) what can you say about \(M\) and \(N\text{?}\) (Refer back to the previous parts of this activity.)
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Answer.
Based on what we have seen in parts (b) and (c), we can conclude that \(M=N\text{.}\)
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Observation 5.5.4.

Notice this fact will eventually help us solve logarithmic equations. If we have an equation where each side is expressed as a single logarithm with matching bases (such as \(\log_b M = \log_bN\)), then it follows that the arguments (\(M\) and \(N\)) are also equal to each other.
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Remark 5.5.5.

Recall that exponential functions and logarithmic functions are inverses. We know that \(\log_b(b^{k})=k\) and according to the law of exponents, we know that:
\begin{equation*} x^a \cdot x^b=x^{a+b} \end{equation*}
\begin{equation*} \dfrac{x^a}{x^b}=x^{a-b} \end{equation*}
\begin{equation*} \left(x^a\right)^b=x^{a \cdot b} \end{equation*}
Consider all these as you move through the activities in this section.
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Activity 5.5.6.

Let’s begin with the law of exponents to see if we can understand the product property of logs. According to the law of exponents, we know that \(10^x \cdot 10^y=10^{x+y}\text{.}\) Start with this equation as you move through this activity.
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(a)
Let \(a=10^x\) and \(b=10^y\text{.}\) How could you rewrite the left side of the equation \(10^x \cdot 10^y\text{?}\)
  1. \(\displaystyle a+b\)
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  2. \(\displaystyle a-b\)
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  3. \(\displaystyle 10^{x+y}\)
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  4. \(\displaystyle a \cdot b\)
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Answer.
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(b)
Recall from Fact 5.5.3 that \(\log_b M=\log_b N\) if and only if \(M=N\text{.}\) Use this property to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?
  1. \(\displaystyle \log_{10}(a+b)=\log_{10}\left(10^{x+y}\right)\)
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  2. \(\displaystyle \log_{10}(a \cdot b)=\log_{10}\left(10^{x+y}\right)\)
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  3. \(\displaystyle \log_{10}(a \cdot b)=\log_{10}\left(10^{a+b}\right)\)
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  4. \(\displaystyle \log_{10}(a+b)=\log_{10}\left(10^{a+b}\right)\)
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Answer.
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(c)
Knowing that \(\log_b(b^k)=k\text{,}\) how could you simplify the right side of the equation?
  1. \(\displaystyle \log_{10}(a+b)\)
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  2. \(\displaystyle \log_{10}(x+y)\)
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  3. \(\displaystyle a+b\)
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  4. \(\displaystyle x+y\)
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Answer.
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(d)
Recall in part (a), we defined \(10^x=a\) and \(10^y=b\text{.}\) What would these look like in logarithmic form?
  1. \(\displaystyle \log_{10}a=x\)
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  2. \(\displaystyle \log_{x}a=10\)
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  3. \(\displaystyle \log_{10}b=y\)
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  4. \(\displaystyle \log_{y}b=10\)
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Answer.
A and C
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(e)
Using your solutions in part (d), how can we rewrite the right side of the equation?
  1. \(\displaystyle 10^{a+b}\)
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  2. \(\displaystyle \log_{10}a-\log_{10}b\)
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  3. \(\displaystyle \log_{10}a+\log_{10}b\)
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  4. \(\displaystyle 10^x+10^y\)
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Answer.
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(f)
Combining parts (a) and (d), which equation represents \(10^x \cdot 10^y=10^{x+y}\) in terms of logarithms?
  1. \(\displaystyle \log_{10}(a+b)=10^{a+b}\)
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  2. \(\displaystyle \log_{10}(a \cdot b)=\log_{10}a-\log_{10}b\)
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  3. \(\displaystyle \log_{10}(a \cdot b)=\log_{10}a+\log_{10}b\)
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  4. \(\displaystyle \log_{10}(a \cdot b)=10^x+10^y\)
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Answer.
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Activity 5.5.7.

According to the law of exponents, we know that \(\dfrac{10^x}{10^y}=10^{x-y}\text{.}\) Start with this equation as you move through this activity.
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(a)
Let \(a=10^x\) and \(b=10^y\text{.}\) How could you rewrite the left side of the equation \(\dfrac{10^x}{10^y}\text{?}\)
  1. \(\displaystyle a+b\)
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  2. \(\displaystyle a-b\)
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  3. \(\displaystyle 10^{x+y}\)
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  4. \(\displaystyle a \cdot b\)
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Answer.
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(b)
Use the one-to-one property of logarithms to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?
  1. \(\displaystyle \log_{10}(a-b)=\log_{10}\left(10^{x-y}\right)\)
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  2. \(\displaystyle \log_{10}\left(\dfrac{a}{b}\right)=\log_{10}\left(10^{x-y}\right)\)
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  3. \(\displaystyle \log_{10}\left(\dfrac{a}{b}\right)=\log_{10}\left(10^{a+b}\right)\)
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  4. \(\displaystyle \log_{10}(a-b)=\log_{10}\left(10^{a-b}\right)\)
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Answer.
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(c)
Knowing that \(\log_b(b^k)=k\text{,}\) how could you simplify the right side of the equation?
  1. \(\displaystyle \log_{10}(a-b)\)
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  2. \(\displaystyle \log_{10}(x-y)\)
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  3. \(\displaystyle x-y\)
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  4. \(\displaystyle a-b\)
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Answer.
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(d)
Recall in part (a), we defined \(10^x=a\) and \(10^y=b\text{.}\) What would these look like in logarithmic form?
  1. \(\displaystyle \log_{10}a=x\)
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  2. \(\displaystyle \log_{x}a=10\)
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  3. \(\displaystyle \log_{10}b=y\)
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  4. \(\displaystyle \log_{y}b=10\)
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Answer.
A and C
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(e)
Using your solutions in part (d), how can we rewrite the right side of the equation?
  1. \(\displaystyle 10^{a+b}\)
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  2. \(\displaystyle \log_{10}a-\log_{10}b\)
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  3. \(\displaystyle \log_{10}a-\log_{10}b\)
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  4. \(\displaystyle 10^{x-y}\)
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Answer.
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(f)
Combining parts (a) and (d), which equation represents \(\dfrac{10^x}{10^y}=10^{x-y}\) in terms of logarithms?
  1. \(\displaystyle \log_{10}(a-b)=10^{a+b}\)
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  2. \(\displaystyle \log_{10}(a-b)=\log_{10}a-\log_{10}b\)
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  3. \(\displaystyle \log_{10}\left(\dfrac{a}{b}\right)=\log_{10}a-\log_{10}b\)
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  4. \(\displaystyle \log_{10}\left(\dfrac{a}{b}\right)=10^{x-y}\)
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Answer.
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Definition 5.5.9.

The product property of logarithms states
\begin{equation*} \log_{a}(m \cdot n)=\log_{a}m+\log_{a}n\text{.} \end{equation*}
The quotient property of logarithms states
\begin{equation*} \log_{a}\left(\dfrac{m}{n}\right)=\log_{a}m-\log_{a}n\text{.} \end{equation*}
Notice that for each of these properties, the base has to be the same.
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Activity 5.5.10.

There is still one more property to consider. This activity will investigate the power property. Suppose you are given
\begin{equation*} \log(x^3)=\log(x \cdot x \cdot x)\text{.} \end{equation*}
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(a)
By applying Definition 5.5.9, how could you rewrite the right-hand side of the equation?
  1. \(\displaystyle \log(3x)\)
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  2. \(\displaystyle \log(x)-\log(x)-\log(x)\)
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  3. \(\displaystyle \log(x^3)\)
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  4. \(\displaystyle \log(x)+\log(x)+\log(x)\)
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Answer.
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(b)
By combining "like" terms, you can simplify the right-hand side of the equation further. What equation do you have after simplifying the right-hand side?
  1. \(\displaystyle \log(x^3)=\log(3x)\)
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  2. \(\displaystyle \log(x^3)=-3\log(x)\)
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  3. \(\displaystyle \log(x^3)=\log(x^3)\)
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  4. \(\displaystyle \log(x^3)=3\log(x)\)
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Answer.
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Definition 5.5.11.

The power property of logarithms states
\begin{equation*} \log_a(m^n)=n \cdot \log_a(m)\text{.} \end{equation*}
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Activity 5.5.12.

Apply Definition 5.5.9 and Definition 5.5.11 to expand the following. (Note: When you are asked to expand logarithmic expressions, your goal is to express a single logarithmic expression into many individual parts or components.)
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(a)
\begin{equation*} \log_3\left(\dfrac{6}{19}\right) \end{equation*}
  1. \(\displaystyle \log_36+\log_319\)
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  2. \(\displaystyle \log_3(6-19)\)
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  3. \(\displaystyle \log_36-\log_319\)
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  4. \(\displaystyle \log_3(6+19)\)
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Answer.
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(b)
\begin{equation*} \log\left((a \cdot b)^2\right) \end{equation*}
  1. \(\displaystyle 2(\log a+\log b)\)
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  2. \(\displaystyle 2\log a-\log b\)
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  3. \(\displaystyle \log a+2\log b\)
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  4. \(\displaystyle 2\log a+2\log b\)
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Answer.
A and D
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(c)
\begin{equation*} \ln\left(\dfrac{x^3}{y}\right) \end{equation*}
  1. \(\displaystyle 3\ln x+\ln y\)
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  2. \(\displaystyle 3\ln x-\ln y\)
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  3. \(\displaystyle 3(\ln x-\ln y)\)
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  4. \(\displaystyle \ln x^3-\ln y\)
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Answer.
B. Note that D is also correct, but it is not expanded fully.
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(d)
\begin{equation*} \log(x \cdot y \cdot z^3) \end{equation*}
  1. \(\displaystyle \log x+\log y+3\log z\)
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  2. \(\displaystyle \log x+\log y+\log z^3\)
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  3. \(\displaystyle \log x-\log y-3\log z\)
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  4. \(\displaystyle 3(\log x+y+z)\)
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Answer.
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Activity 5.5.13.

Apply Definition 5.5.9 and Definition 5.5.11 to condense into a single logarithm.
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(a)
\begin{equation*} 6\log_6 a+3\log_6 b \end{equation*}
  1. \(\displaystyle 6(\log_6 a)+3(\log_6 b)\)
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  2. \(\displaystyle \log_6 a^6+\log_6 b^3\)
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  3. \(\displaystyle (\log_6 a)^6+(\log_6 b)^3\)
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  4. \(\displaystyle \log_6\left(a^6 \cdot b^3\right)\)
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Answer.
D. B is also correct, but it is not condensed fully.
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(b)
\begin{equation*} \ln x-4\ln y \end{equation*}
  1. \(\displaystyle \ln x-\ln y^4\)
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  2. \(\displaystyle \ln\left(\dfrac{x}{y^4}\right)\)
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  3. \(\displaystyle \ln\left(\dfrac{x}{y}\right)^4\)
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  4. \(\displaystyle \ln(x \cdot y^4)\)
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Answer.
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(c)
\begin{equation*} 2(\log(2x)-\log y) \end{equation*}
  1. \(\displaystyle \log\left(\dfrac{4x^2}{y^2}\right)\)
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  2. \(\displaystyle \log\left(\dfrac{2x^2}{y^2}\right)\)
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  3. \(\displaystyle 2 \cdot \log\left(\dfrac{2x}{y}\right)\)
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  4. \(\displaystyle \log\left(\dfrac{4x^2}{y}\right)\)
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Answer.
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(d)
\begin{equation*} \log 3+2\log 5 \end{equation*}
  1. \(\displaystyle \log(3 \cdot 5^2)\)
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  2. \(\displaystyle 2 \cdot \log 75\)
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  3. \(\displaystyle \log 75\)
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  4. \(\displaystyle \log 225\)
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Answer.
C. Note that A is also correct, but not condensed completely.
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Remark 5.5.14.

You might have noticed that a scientific calculator has only "log" and "ln" buttons (because those are the most common bases we use), but not all logs have base 10 or e as their bases.
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Activity 5.5.15.

Suppose you wanted to find the value of \(\log_53\) in your calculator but you do not know how to input a base other than \(10\) or \(e\) (i.e., you only have the "log" and "ln" buttons on your calculator). Let’s explore another helpful tool that can help us find the value of \(\log_53\text{.}\)
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(a)
Let’s start with the general statement, \(\log_ba=x\text{.}\) How can we rewrite this logarithmic equation into an exponential equation?
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Answer.
\(b^x=a\)
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(b)
Now take the log of both sides of your equation and apply the power property of logarithms to bring the exponent down. What equation do you have now?
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Answer.
\(x \cdot \log b=\log a\)
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(c)
Solve for \(x\text{.}\) What does \(x\) equal?
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Answer.
\(x=\dfrac{\log a}{\log b}\)
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(d)
Recall that when we started, we defined \(x=\log_ba\text{.}\) Substitute \(\log_ba\) into your equation you got in part c for \(x\text{.}\) What is the resulting equation?
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Answer.
\(\log_ba=\dfrac{\log a}{\log b}\)
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(e)
Apply what you got in part d to find the value of \(\log_53\text{.}\) What is the approximate value of \(\log_53\text{?}\)
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Answer.
\(\log_53\) is approximately \(0.683\)
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Remark 5.5.16.

Notice in Activity 5.5.15, we were able to calculate \(\log_53\) using logs of base \(10\text{.}\) You should now be able to find the value of a logarithm of any base!
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Definition 5.5.17.

The change of base formula is used to write a logarithm of a number with a given base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm.
\begin{equation*} \log_{b}a=\dfrac{\log a}{\log b} \end{equation*}
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Activity 5.5.18.

Apply Definition 5.5.17 and a calculator to approximate the value of each logarithm.
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(a)
\(\log_2 30\)
  1. \(\displaystyle 0.204\)
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  2. \(\displaystyle 4.907\)
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  3. \(\displaystyle \dfrac{\log30}{\log2}\)
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  4. \(\displaystyle \dfrac{\log2}{\log30}\)
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Answer.
B and C
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(b)
\(\ln 183\)
  1. \(\displaystyle \dfrac{\ln183}{\ln e}\)
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  2. \(\displaystyle 67.32\)
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  3. \(\displaystyle \dfrac{\ln e}{\ln183}\)
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  4. \(\displaystyle 5.209\)
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Answer.
A and D
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Subsection 5.5.2 Exercises

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