12.4: Logarithm properties
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In this section, we take logarithms one step further and discuss properties of logarithms. Since logarithms are exponents, and we have many exponent properties as we learned in the Polynomials chapter, it makes sense we have similar properties for logarithms. E.g., if the product of two factors with the same base result in the sum of their exponents, then we have the product property of logarithms; if the quotient of two factors with the same base result in the difference of their exponents, then we have the quotient property of logarithms; a similar case for the power rule of logarithms.
Understand Properties of Logarithms
A logarithm of a product is the sum of the logarithms:
\[\log_a (MN) = \log_a M + \log_a N\nonumber\]
where \(a\) is the base, \(a > 0\) and \(a\neq 1\), and \(M,\: N > 0\).
Rewrite as a sum of logarithms: \(\log_3 (6\cdot 5)\)
Solution
Since \(3\) is the base and \(6\) and \(5\) are the factors, we see in the formula \(\log_a (MN)\), \(a = 3\), \(M = 6\), and \(N = 5\). Hence,
\[\log_3 (6\cdot 5)=\log_3 6+\log_3 5\nonumber\]
Rewrite as a sum of logarithms: \(\ln(2k)\)
Solution
Since \(e\) is the base and \(2\) and \(k\) are the factors (you see this when we write \(2k\) as \(2\cdot k\)), we see in the formula \(\log_a (MN)\), \(a = e\), \(M = 2\), and \(N = k\). Hence,
\[\ln(2k) = \log_e (2\cdot k) = \log_e 2 + \log_e k = \ln 2 + \ln k\nonumber\]
A logarithm of a quotient is the difference of the logarithms:
\[\log_a \left(\dfrac{M}{N}\right) = \log_a M − \log_a N\nonumber\]
where \(a\) is the base, \(a > 0\) and \(a\neq 1\), and \(M,\: N > 0\).
Rewrite as a difference of logarithms: \(\log_3\left(\dfrac{7}{5}\right)\)
Solution
Since \(3\) is the base, \(7\) is the numerator, and \(5\) is the denominator, we see in the formula \(\log_a\left(\dfrac{M}{N}\right)\), \(a = 3\), \(M = 7\), and \(N = 5\). Hence,
\[\log_3\left(\dfrac{7}{5}\right)=\log_3 7-\log_3 5\nonumber\]
Be careful to observe that the value of the log after the minus sign is the value of the denominator of the fraction.
Rewrite as a difference of logarithms: \(\ln\left(\dfrac{7}{2}\right)\)
Solution
Since \(e\) is the base, \(7\) is the numerator, and \(2\) is the denominator, we see in the formula \(\log_a\left(\dfrac{M}{N}\right)\), \(a = e\), \(M = 7\), and \(N = 2\). Hence,
\[\ln\left(\dfrac{7}{2}\right)=\log_e\left(\dfrac{7}{2}\right)=\log_e 7-\log_e 2=\ln 7-\ln 2\nonumber\]
A logarithm of a power is the product of the power and logarithm:
\[\log_a M^p =p\log_a M\nonumber\]
where \(a\) is the base, \(a>0\) and \(a\neq 1\), and \(M>0\).
Rewrite all powers as factors: \(\log_7 2^4\).
Solution
Since \(4\) is the power on \(2\), then we can bring down \(4\) in front of the log:
\[\begin{aligned}\log_7 2^{\color{blue}{4}}& \color{black}{=}\color{blue}{4}\color{black}{\cdot}\log_7 2 \\ &=4\log_7 2\end{aligned}\]
Notice \(4\) and \(\log_7 2\) become factors.
Rewrite all powers as factors: \(\ln x^{\sqrt{2}}\).
Solution
Since \(\sqrt{2}\) is the power on \(x\), then we can bring down \(\sqrt{2}\) in front of the \(\ln\):
\[\begin{aligned}\ln x^{\color{blue}{\sqrt{2}}}&\color{black}{=}\color{blue}{\sqrt{2}}\color{black}{\cdot}\ln x \\ &=\sqrt{2}\ln x\end{aligned}\]
Notice \(\sqrt{2}\) and \(\ln x\) become factors.
Other Properties of Logarithms
Here are a few other properties of logarithms that we find useful when simplifying. Recall, we use these properties to have better technique when we have to solve equations with logarithms.
If \(a\), \(M>0\), and \(a\neq 1\), then
\[\log_a 1=0\quad \log_a a=1\nonumber\]
\[a^{\log_a M}=M\quad \log_a a^r=r\nonumber\]
Evaluate each logarithm.
- \(\log_5 1\)
- \(\log 10\)
- \(\log 10^{-4}\)
- \(12^{\log_{12}\sqrt{12}}\)
Solution
- Since we need to find \(5^? = 1\), then by the first property we know the result is zero. Thus, \(\log_5 1 = 0\).
- First, the log has no visible base. By default, we use the common logarithm and assume the base is \(10\). So, since we need to find \(10^? = 10\), then by the second property we know the result is one. Thus, \(\log 10 = 1\).
- First, the log has no visible base. By default, we use the common logarithm and assume the base is \(10\). So, since we need to find \(10^? = 10^{−4}\), then by the last property we know the result is \(−4\). Thus, \(\log 10^{−4} = −4\).
- If we rewrite this in logarithmic form, we get \[\log_{12}?=\log_{12}\sqrt{12}\nonumber\] We can easily see if this statement has to be true, then \(? =\sqrt{12}\). Also, by the third property, we know the result is \(\sqrt{12}\). Thus, \(12^{\log_{12}\sqrt{12}}=\sqrt{12}\).
Expand and Contract Logarithms
We discuss expanding and contracting logarithmic expressions as part of applying the properties. In a future section, we apply these properties to solve logarithmic equations.
When expanding logarithms from a single expression, be sure to write all logarithms of
Rule 1. Products as sums
Rule 2. Quotients as differences
Rule 3. Powers as factors
We use order of operations when expanding an expression and apply the power property, and then product and quotient properties- in that order.
Expand the logarithm by rewriting as a sum or difference of logarithms with powers as factors.
\[\log\left(\dfrac{1000\sqrt{x}}{y}\right)\nonumber\]
Solution
We see a quotient for the value of the logarithm, so we foresee we will use the quotient property of logarithms. If we look closer at the numerator, we see there is a product of two factors. Hence, we will use the product property of logarithms, too. Furthermore, we will have to use the power property of logarithms.
\[\begin{array}{rl} \log\left(\dfrac{1000\sqrt{x}}{y}\right) &\text{Apply quotient property of logarithms} \\ \log(1000\cdot\sqrt{x})-\log y&\text{Apply product property of logarithms} \\ \log 1000 +\log(\sqrt{x})-\log y&\text{Rewrite }\sqrt{x}\text{ as }x^{1/2} \\ \log 1000+\log x^{1/2}-\log y &\text{Apply power property of logarithms} \\ \log 1000+\dfrac{1}{2}\log x-\log y &\text{Expanded logarithmic expression}\end{array}\nonumber\]
Notice, we had to rewrite \(\sqrt{x}\) as \(x^{1/2}\) in order to see there was a power on \(x\) in which we had to use the product property of logarithms to bring it down as a factor. Thus, all products are written as sums, all quotients are written as differences, and all powers are written as factor
When contracting logarithms from a single expression, be sure to write any
Rule 1. Multiple of a logarithm as a power of the argument
Rule 2. Sums of logarithms as a logarithm of a product
Rule 3. Differences of logarithms as a logarithm of a quotient
Write \(\log_2 9 + 2 \log_2 x − \log_2 (x − 4)\) as a single logarithm
Solution
Right away, we see a sum and difference with logarithms, so we know we will use the quotient and product property of logarithms. Furthermore, we will have to use the power property of logarithms.
\[\begin{array}{rl} \log_2 9+\color{red}{2}\color{black}{\log_2}x-\log_2(x-4)&\text{Apply power property of logarithms} \\ \color{blue}{\log_2 9+\log_2 x}\color{red}{^{2}}\color{black}{-}\log_2(x-4)&\text{Apply product property of logarithms} \\ \color{blue}{\log_2 9x^2}\color{black}{-}\log_2 (x-4)&\text{Apply quotient property of logarithms} \\ \log_2\left(\dfrac{9x^2}{x-4}\right)&\text{Contracted logarithmic expression}\end{array}\nonumber\]
Notice, we had to rewrite \(2 \log_2 x\) as \(\log_2 x^2\) in order to see there was a power on \(x\) in which we had to use the product property of logarithms to write \(2\) as the exponent. Thus, all factors are written as powers, all sums are written as products, and all differences are written as quotients.
The Scottish mathematician John Napier published his discovery of logarithms in 1614. His purpose was to assist in the multiplication of quantities that were then called sines. The whole sine was the value of the side of a right-angled triangle with a large hypotenuse.
Change of Base Formula
la Sometimes we need to be able to rewrite logarithms in terms of other bases. This is especially helpful when counting in different numeration systems. For example, in the computer language, we count in a binary numeration system, base \(2\). We can use the change of base formula to rewrite numbers in different bases and it is particularly useful in computer science. However, in this textbook, we learn the change of base formula for the common and natural logarithm bases, i.e., base \(10\) and base \(e\).
Let’s take a simple general exponential equation \(a^y = M\). We usually rewrite its logarithmic form as \(y = \log_a M\). Well, now, let’s solve for \(y\) by taking the common logarithm, log, to each side:
\[\begin{array}{rl} a^y=M &\text{Take common logarithm to each side} \\ \log a^y =\log M&\text{Apply the power rule of logarithms} \\ y\log a=\log M&\text{Solve for }y \\ y=\dfrac{\log M}{\log a}&\text{This is the change of base formula}\end{array}\nonumber\]
If \(a\), \(b\), \(M>0\), and \(a\), \(b\neq 1\), then
\[\log_a M=\dfrac{\log M}{\log a}\quad\text{or}\quad \log_a M=\dfrac{\ln M}{\ln a}\nonumber\]
where log is the common logarithm, and ln is the natural logarithm. We can either formula and obtain the same result.
Rewrite the expression using the Change of Base formula and then approximate the answer to three decimal places.
\[\log_2 9\nonumber\]
Solution
We would like to approximate this value using a calculator, but we cannot easily enter a logarithm in base \(2\). We must rewrite \(\log_2 9\) so that we can easily enter it into the calculator. This is where the Change of Base (COB) formula comes in handy. Notice the base \(a = 2\) and the value \(M = 9\). Using the COB formula, we rewrite \(\log_2 9\) as
\[\log_{\color{red}{2}}\color{blue}{9}\color{black}{=}\dfrac{\log\color{blue}{9}}{\log\color{red}{2}}\nonumber\]
Recall, log is the common logarithm, \(\log_{10}\). Putting \(\dfrac{\log 9}{\log 2}\) into the calculator, we approximate \(3.170\).
We could have easily used the natural logarithm in the COB formula and would have obtained the same result. There’s no need to use both formulas- one will suffice.
Logarithm Properties Homework
Write the expression as a logarithm of a single expression. Assume that variables represent positive numbers.
\(\log_a m-\log_a n+6\log_a k\)
\(\log_8 6+\log_8 x\)
\(\log_8 3+\log_8 (x^{3}-2)+\log_8 2\)
\(3\log_a (2x+1)-2\log_a(2x-1)+2\)
Write as the sum and/or difference of logarithms. Express powers as factors.
\(\log_{4}\left(\dfrac{64}{\sqrt{x-1}}\right)\)
\(\log_2\left(\dfrac{x^2}{y^6}\right)\)
\(\log_b (xz^3)\)
\(\log_b\left(\dfrac{xy^5}{z^7}\right)\)
Use the Change of Base Formula and a calculator to evaluate the logarithm. Round to four decimal places.
\(\log_3 23\)
\(\log_{0.4}20\)
\(\log_{19}57.8\)
Evaluate each logarithm.
\(\log_{23}23\)
\(\log_{\sqrt{11}}\left(\sqrt{11}^{0.394}\right)\)
\(247^{\log_{247}\sqrt{5}}\)
\(\log_{\dfrac{1}{3}}1\)