18.2.2: Properties of Logarithmic Functions
- Page ID
- 76928
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- Express the logarithm of a quotient as a difference.
- Express the logarithm of a power as a product.
- Simplify logarithmic expressions.
Introduction
Throughout your study of algebra, you have come across many properties such as the commutative, associative, and distributive properties. These properties help you take a complicated expression or equation and simplify it.
The same is true with logarithms. There are a number of properties that will help you simplify complex logarithmic expressions. Since logarithms are so closely related to exponential expressions, it is not surprising that the properties of logarithms are very similar to the properties of exponents. As a quick refresher, here are the exponent properties.
Product of powers: \(b^{m} b^{n}=b^{m+n}\)
Quotient of powers: \(\frac{b^{m}}{b^{n}}=b^{m-n}\)
Power of a power: \(\left(b^{m}\right)^{n}=b^{m n}\)
One important but basic property of logarithms is \(\log _{b} b^{x}=x\). This makes sense when you convert the statement to the equivalent exponential equation. The result? \(b^{x}=b^{x}\).
Let’s find the value of \(y\) in \(\log _{3} 3^{2}=y\). Remember \(\log _{b} x=y \Leftrightarrow b^{y}=x\), so \(\log _{3} 3^{2}=y\) means \(3^{y}=3^{2}\) and \(y\) must be 2, which means \(\log _{3} 3^{2}=2\). You will get the same answer that \(\log _{3} 3^{2}\) equals 2 by using the property that \(\log _{b} b^{x}=x\).
Logarithm of a Product
Remember that the properties of exponents and logarithms are very similar. With exponents, to multiply two numbers with the same base, you add the exponents. With logarithms, the logarithm of a product is the sum of the logarithms.
The logarithm of a product is the sum of the logarithms: \(\log _{b}(M N)=\log _{b} M+\log _{b} N\)
Let’s try the following example.
Problem: Use the product property to rewrite \(\log _{2}(4 \cdot 8)\).
Answer
Use the product property to write as a sum.
\(\log _{2}(4 \cdot 8)=\log _{2} 4+\log _{2} 8\)
Simplify each addend, if possible. In this case, you can simplify both addends. First rewrite \(\log _{2} 4\) as \(\log _{2} 2^{2}\) and \(\log _{2} 8\) as \(\log _{2} 2^{3}\), and then use the property \(\log _{b} b^{x}=x\).
Or, rewrite \(\log _{2} 4=y\) as \(2^{y}=4\) to find \(y = 2\), and \(\log _{2} 8=y\) as \(2^{y}=8\) to find \(y = 3\).
Use whatever method makes sense to you.
\(\begin{aligned}
\log _{2} 4+\log _{2} 8 &=\\
\log _{2} 2^{2}+\log _{2} 2^{3} &=\\
2+3 &=5
\end{aligned}\)
\(\log _{2}(4 \cdot 8)=5\)
Another way to simplify \(\log _{2}(4 \cdot 8)\) would be to multiply 4 and 8 as a first step.
\(\log _{2}(4 \cdot 8)=\log _{2} 32=5 \text { because } 2^{5}=32\)
You get the same answer\(\log _{2}(4 \cdot 8) = 5\) as in the example!
Notice the similarity to the exponent property: \(b^{m} b^{n}=b^{m+n}\), while \(\log _{b}(M N)=\log _{b} M+\log _{b} N\). In both cases, a product becomes a sum.
Problem: Use the product property to rewrite \(\log _{3}(9 x)\).
Answer
Use the product property to write as a sum.
\(\log _{3}(9 x)=\log _{3} 9+\log _{3} x\)
Simplify each addend, if possible. In this case, you can simplify \(\log _{3} 9\) but not \(\log _{3} x\). Rewrite \(\log _{3} 9\ as \(\log _{3} 3^{2}\) and then use the property \(\log _{b} b^{x}=x\).
Or, simplify \(\log _{3} 9\) by converting \(\log _{3} 9=y\) to \(3^{y}=9\) and finding that \(y = 2\). Use whatever method makes sense to you.
\(\log _{3} 9+\log _{3} x=\log _{3} 3^{2}+\log _{3} x=2+\log _{3} x\)
\(\log _{3}(9 x)=2+\log _{3} x\)
If the product has many factors, you just add the individual logarithms:
\(\log _{b}(A B C D)=\log _{b} A+\log _{b} B+\log _{b} C+\log _{b} D\)
Rewrite \(\log _{2} 8 a\), then simplify.
- \(3 \log _{2} a\)
- \(\log _{2} 3 a\)
- \(\log _{2}(3+a)\)
- \(3+\log _{2} a\)
- Answer
-
- Incorrect. The individual logarithms must be added, not multiplied. The correct answer is \(3+\log _{2} a\).
- Incorrect. You found that \(\log _{2} 8=3\), but you must first apply the logarithm of a product property. The correct answer is \(3+\log _{2} a\).
- Incorrect. The logarithm of a product property says you separate the 8 and \(a\) into separate logarithms. The correct answer is \(3+\log _{2} a\).
- Correct. The logarithm of a product property says \(\log _{2} 8 a=\log _{2} 8+\log _{2} a\), and \(\log _{2} 8=3\).
Logarithm of a Quotient
You can use the similarity between the properties of exponents and logarithms to find the property for the logarithm of a quotient. With exponents, to multiply two numbers with the same base, you add the exponents. To divide two numbers with the same base, you subtract the exponents. What do you think the property for the logarithm of a quotient will look like?
As you may have suspected, the logarithm of a quotient is the difference of the logarithms.
\(\log _{b}\left(\frac{M}{N}\right)=\log _{b} M-\log _{b} N\)
With both properties: \(\frac{b^{m}}{b^{n}}=b^{m-n}\) and \(\log _{b}\left(\frac{M}{N}\right)=\log _{b} M-\log _{b} N\), a quotient becomes a difference.
Problem: Use the quotient property to rewrite \(\log _{x}\left(\frac{x}{2}\right)\).
Answer
Use the quotient property to rewrite as a difference.
\(\log _{2}\left(\frac{x}{2}\right)=\log _{2} x-\log _{2} 2\)
The first expression can’t be simplified further. However, the second expression can be simplified. What exponent on the base (2) gives a result of 2? Since \(2^1 = 2\), you know \(\log _{2} 2=1\).
\(\log _{2}\left(\frac{x}{2}\right)=\log _{2} x-1\)
Which of these is equivalent to \(\log _{3}\left(\frac{81}{a}\right)\)?
- \(4-\log _{3} a\)
- \(\frac{4}{\log _{3} a}\)
- \(\log _{3}(4-a)\)
- \(\log _{3}\left(\frac{4}{a}\right)\)
- Answer
-
- Correct. The logarithm of a quotient property states \(\log _{3}\left(\frac{81}{a}\right)=\log _{3} 81-\log _{3} a\), and \(\log _{3} 81=4\).
- Incorrect. The individual logarithms must be subtracted, not divided. The correct answer is \(4-\log _{3} a\).
- Incorrect. The logarithm of a quotient property says you separate the 81 and \(a\) into separate logarithms. The correct answer is \(4-\log _{3} a\).
- Incorrect. You found that \(\log _{3} 81=4\), but you must first apply the logarithm of the quotient property. The correct answer is \(4-\log _{3} a\).
Logarithm of a Power
The remaining exponent property was power of a power: \(\left(b^{m}\right)^{n}=b^{m n}\). The similarity with the logarithm of a power is a little harder to see.
\(\log _{b} M^{n}=n \log _{b} M\)
With both properties, \(\left(b^{m}\right)^{n}=b^{m n}\) and \(\log _{b} M^{n}=n \log _{b} M\), the power “\(n\)” becomes a factor.
Problem: Use the power property to simplify \(\log _{3} 9^{4}\).
Answer
You could find \(9^4\) but that wouldn’t make it easier to simplify the logarithm. Use the power property to rewrite \(\log _{3} 9^{4}\) as \(4 \log _{3} 9\).
\(\log _{3} 9^{4}=4 \log _{3} 9\)
You may be able to recognize by now that since \(3^2 = 9\), \(\log _{3} 9=2\).
\(4 \log _{3} 9=4 \cdot 2\)
Multiply the factors.
\(\log _{3} 9^{4}=8\)
Notice in this case that you also could have simplified it by rewriting it as 3 to a power: \(\log _{3} 9^{4}=\log _{3}\left(3^{2}\right)^{4}\). Using exponent properties, this is \(\log _{3} 3^{8}\) and by the property \(\log _{b} b^{x}=x\), this must be 8!
Problem: Use the properties of logarithms to rewrite \(\log _{4} 64^{x}\).
Answer
Use the power property to rewrite \(\log _{4} 64^{x}\) as \(x \log _{4} 64\).
\(64=4 \cdot 4 \cdot 4=4^{3}\)
Rewrite \(\log _{4} 64\) as \(\log _{4} 4^{3}\), then use the property \(\log _{b} b^{x}=x\) to simplify \(\log _{4} 4^{3}\).
Or, you may be able to recognize by now that since \(4^{3}=64\), \(\log _{4} 64=3\).
\(\begin{array}{c}
\log _{4} 64^{x}=x \log _{4} 64 \\
=x \log _{4} 4^{3} \\
\quad=x \cdot 3
\end{array}\)
Multiply the factors.
\(\log _{4} 64^{x}=3 x\)
Which of these is equivalent to \(\log _{2} x^{8}\)?
- \(\log _{2} 3 x\)
- \(8 \log _{2} x\)
- \(\log _{2} 8 x\)
- \(3 \log _{2} x\)
- Answer
-
- Incorrect. The exponent becomes a factor outside the logarithm. The correct answer is \(8 \log _{2} x\).
- Correct. By the power property, \(\log _{2} x^{8}=8 \log _{2} x\). You can’t simplify this further.
- Incorrect. The exponent becomes a factor outside the logarithm. The correct answer is \(8 \log _{2} x\).
- Incorrect. You probably noticed that \(\log _{2} 8=3\), so you used 3 instead of 8 when you pulled the exponent out to be a factor. However, the exponent must be pulled outside the logarithm to be a factor without any other changes. The correct answer is \(8 \log _{2} x\).
Simplifying Logarithmic Expressions
The properties can be combined to simplify more complicated expressions involving logarithms.
Problem: Use the properties of logarithms to expand \(\log _{10}\left(\frac{a b}{c d}\right)\) into four simpler terms.
Answer
Use the quotient property to rewrite \(\log _{10}\left(\frac{a b}{c d}\right)\) as a difference of logarithms.
\(\log _{10}\left(\frac{a b}{c d}\right)=\log _{10}(a b)-\log _{10}(c d)\)
Now you have two logarithms, each with a product. Apply the product rule to each.
Be careful with the subtraction! Since all of \(\log _{10} c d\) is subtracted, you have to subtract both parts of the term: \(\left(\log _{10} c+\log _{10} d\right)\)
\(\log _{10}(a b)-\log _{10}(c d)=\log _{10} a+\log _{10} b-\left(\log _{10} c+\log _{10} d\right)\)
\(\log _{10}\left(\frac{a b}{c d}\right)=\log _{10} a+\log _{10} b-\log _{10} c-\log _{10} d\)
Problem: Simplify \(\log _{6}(a b)^{4}\), writing it as two separate terms.
Answer
Use the power property to rewrite \(\log _{6}(a b)^{4}\) as \(4 \log _{6}(a b)\).
You are taking the log of a product, so apply the product property.
Be careful: the value 4 is multiplied by the whole logarithm, so use parentheses when you rewrite \(\log _{6}(a b)\) as \(\left(\log _{6} a+\log _{6} b\right)\).
\(\begin{array}{l}
\log _{6}(a b)^{4}=4 \log _{6}(a b) \\
\quad=4\left(\log _{b} a+\log _{6} b\right)
\end{array}\)
Use the distributive property.
\(\log _{6}(a b)^{4}=4 \log _{6} a+4 \log _{6} b\)
Simplify \(\log _{3} x^{2} y\).
- \(2\left(\log _{3} x+\log _{3} y\right)\)
- \(\log _{3} x^{2}+\log _{3} y\)
- \(2 \log _{3} x y\)
- \(2 \log _{3} x+\log _{3} y\)
- Answer
-
- Incorrect. You may have started incorrectly by applying the power property, or you may have started correctly with the product property but then incorrectly applied the power property. The correct answer is \(2 \log _{3} x+\log _{3} y\).
- Incorrect. While you correctly applied the product property first, \(\log _{3} x^{2}\) can be simplified further. The correct answer is \(2 \log _{3} x+\log _{3} y\).
- Incorrect. You probably started incorrectly by applying the power property. Start with the product property. The correct answer is \(2 \log _{3} x+\log _{3} y\).
- Correct. \(\log _{3} x^{2} y=\log _{3} x^{2}+\log _{3} y=2 \log _{3} x+\log _{3} y\)
Summary
Like exponents, logarithms have properties that allow you to simplify logarithms when their inputs are a product, a quotient, or a value taken to a power. The properties of exponents and the properties of logarithms have similar forms.
| Exponents | Logarithms | |
|---|---|---|
| Product Property | \(b^{m} b^{n}=b^{m+n}\) | \(\log _{b}(M N)=\log _{b} M+\log _{b} N\) |
| Quotient Property | \(\frac{b^{m}}{b^{n}}=b^{m-n}\) | \(\log _{b}\left(\frac{M}{N}\right)=\log _{b} M-\log _{b} N\) |
| Power Property | \(\left(b^{m}\right)^{n}=b^{m n}\) | \(\log _{b} M^{n}=n \log _{b} M\) |
Notice how the product property leads to addition, the quotient property leads to subtraction, and the power property leads to multiplication for both exponents and logarithms.