4.3.3: Properties of Logarithms
By the end of this section, you will be able to:
- Use the properties of logarithms
- Use the Change of Base Formula
Before you get started, take this readiness quiz.
1. Evaluate
a. \(b^{0}\)
b. \(b^{1}\)
2. Write \(\sqrt[3]{x^{2} y}\) with a rational exponent.
3. Round \(2.5646415\) to three decimal places.
Use the Properties of Logarithms
Now that we have learned about exponential and logarithmic expressions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.
The first two properties derive from the definition of logarithms. Since \(b^{0}=1\), we can convert this to logarithmic form and get \(\log _{b} (1)=0\). Also, since \(b^{1}=b\), we get \(\log _{b} (b)=1\).
\(\log _{b} (1)=0\)
\(\log _{b} (b)=1\)
In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.
Evaluate using the properties of logarithms:
a. \(\log _{8} (1)\)
b. \(\log _{6} (6)\)
Solution
a.
| \(\quad \log _{8} (1)\) | |
|---|---|
| Use the property, \(\log_{b} (1)=0\). | \(=0\) |
b.
| \(\quad \log _{6} (6)\) | |
|---|---|
| Use the property, \(\log_{b} (b)=1\). | \(=1\) |
Evaluate using the properties of logarithms:
a. \(\log _{13} (1)\)
b. \(\log _{9} (9)\)
- Answer
-
a. \(0\)
b. \(1\)
Evaluate using the properties of logarithms:
a. \(\log _{5} (1)\)
b. \(\log _{7} (7)\)
- Answer
-
a. \(0\)
b. \(1\)
The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.
The exponential equation \(b^{\log _{b} (x)}=x\) converts to the logarithmic equation \(\log _{b} (x)=\log _{b} (x)\), which is a true statement for positive values for \(x\) only.
The logarithmic equation \(\log _{b} \left(b^{x}\right)=x\) converts to the exponential equation \(b^{x}=b^{x}\), which is also a true statement (for all values of \(x\)).
These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power.
For \(b>0\) and \(b \neq 1\),
\(b^{\log _{b} (x)}=x \) for \(x>0\) and \(\log _{b} \left(b^{x}\right)=x.\)
In the next example, we apply the inverse properties of logarithms.
Evaluate using the properties of logarithms:
a. \(4^{\log _{4} (9)}\)
b. \(\log _{3} \left(3^{5}\right)\)
Solution
a.
| \(\quad 4^{\log _{4}(9)}\) | |
|---|---|
| Use the property, \(b^{\log _{b} (x)}=x\). | \(=9 \) |
b.
| \(\quad \log _{3} \left(3^{5}\right)\) | |
|---|---|
| Use the property, \(\log _{b} \left(b^{x}\right)=x\). | \(=5\) |
Evaluate using the properties of logarithms:
a. \(5^{\log _{5} \left(15\right)}\)
b. \(\log _{7} \left(7^{4}\right)\)
- Answer
-
a. \(15\)
b. \(4\)
Evaluate using the properties of logarithms:
a. \(2^{\log _{2} \left(8\right)}\)
b. \(\log _{2} \left(2^{15}\right)\)
- Answer
-
a. \(8\)
b. \(15\)
There are three more properties of logarithms that will be useful in our work. We know exponential expressions and logarithmic expressions are very interrelated. Our definition of logarithm shows us that a logarithm is the exponent of the equivalent exponential. The properties of exponents have related properties for exponents.
In the Product Property of Exponents, \(b^{m} b^{n}=b^{m+n}\), we see that to multiply the same base, we add the exponents. The Product Property of Logarithms , \(\log_{b} (MN)=\log_{b} (M)+\log_{b} (N)\), tells us that to take the log of a product, we add the log of the factors (here we use \(M=b^{m}\) and \(N= b^{n}\) and note that \(\log_b \left(b^{m+n}\right)=m+n=\log_b \left(b^m\right)+\log_b \left(b^n\right)\).
If \(M>0\), \(N>0\), \(b>0\) and \(b\neq 1,\) then
\(\log _{b}(MN)=\log _{b} (M)+\log _{b} (N).\)
The logarithm of a product is the sum of the logarithms.
We use this property to write the log of a product as a sum of the logs of each factor.
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible:
a. \(\log _{3} (7 x)\)
b. \(\log _{4} (64 x y)\)
Solution
a.
| \(\quad \log _{3} (7 x)\) | |
|---|---|
| Use the Product Property, \(\log _{b}(M N)=\log_{b} (M)+\log_{b} (N)\). | \(=\log _{3} (7)+\log _{3} (x)\) |
b.
| \(\quad\log _{4}(64 x y)\) | |
|---|---|
| Use the Product Property, \(\log _{b}(M N)=\log _{b} (M)+\log _{b} (N)\). | \(=\log _{4} (64)+\log _{4} (x)+\log _{4} (y)\) |
| Simplify by evaluating \(\log _{4} (64)\). | \(=3+\log _{4} (x)+\log _{4} (y)\) |
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.
a. \(\log _{3} (3 x)\)
b. \(\log _{2} (8 x y)\)
- Answer
-
a. \(1+\log _{3} (x)\)
b. \(3+\log _{2} (x)+\log _{2} (y)\)
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.
a. \(\log _{9} (9 x)\)
b. \(\log _{3} (27 x y)\)
- Answer
-
a. \(1+\log _{9} (x)\)
b. \(3+\log _{3} (x)+\log _{3} (y)\)
Similarly, in the Quotient Property of Exponents, \(\dfrac{b^{m}}{b^{n}}=b^{m-n}\), we see that to divide the same base, we subtract the exponents. The Quotient Property of Logarithms , \(\log_{b} \left(\dfrac{M}{N}\right)=\log_{b} (M)-\log_{b} (N)\) tells us to take the log of a quotient, we subtract the log of the numerator and denominator.
If \(M>0\), \(N>0\), \(b>0\) and \(b \neq 1,\) then
\(\log_{b} \left(\dfrac{M}{N}\right)=\log_{b} (M)-\log_{b} (N).\)
The logarithm of a quotient is the difference of the logarithms.
Note that \(\log_{b} (M)-\log_{b} (N) \not=\log_{b}(M-N)!\).
We use this property to write the log of a quotient as a difference of the logs of each factor.
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
a. \(\log _{5} \left(\dfrac{5}{7}\right)\)
b. \(\log \left(\dfrac{x}{100}\right)\)
Solution
- a.
-
\(\quad\log _{5} \left(\dfrac{5}{7}\right)\) Use the Quotient Property, \(\log_{b} \left(\dfrac{M}{N}\right)=\log_{b} (M)-\log_{b} (N)\). \(=\log _{5} (5)-\log _{5} (7)\) Simplify. \(=1-\log _{5} (7)\) b.
\(\quad\log \left(\dfrac{x}{100}\right)\) Use the Quotient Property, Use the Quotient Property, \(\log_{b} \left(\dfrac{M}{N}\right)=\log_{b} (M)-\log_{b} (N)\). \(=\log (x)-\log (100)\) Simplify. \(=\log (x)-2\)
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
a. \(\log _{4} \left(\dfrac{3}{4}\right)\)
b. \(\log \left(\dfrac{x}{1000}\right)\)
- Answer
-
a. \(\log _{4} (3)-1\)
b. \(\log (x)-3\)
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
a. \(\log _{2} \left(\dfrac{5}{4}\right)\)
b. \(\log \left(\dfrac{10}{y}\right)\)
- Answer
-
a. \(\log _{2} (5)-2\)
b. \(1-\log (y)\)
The third property of logarithms is related to the Power Property of Exponents, \(\left(a^{m}\right)^{n}=a^{m \cdot n}\), we see that to raise a power to a power, we multiply the exponents. The Power Property of Logarithms , \(\log_{b} M^{p}=p \log_{b} M\) tells us to take the log of a number raised to a power, we multiply the power times the log of the number.
If \(M>0\), \(b>0\), \(b \neq 1\) and \(p\) is any real number then,
\(\log_{b} \left(M^{p}\right)=p \log_{b} (M).\)
The log of a number raised to a power as the product product of the power times the log of the number.
We use this property to write the log of a number raised to a power as the product of the power times the log of the number. We essentially take the exponent and throw it in front of the logarithm.
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
a. \(\log _{5} \left(4^{3}\right)\)
b. \(\log \left(x^{10}\right)\)
Solution
a.
| \(\quad\log _{5} \left(4^{3}\right)\) | |
|---|---|
| Use the Power Property, \(\log_{b}\left( M^{p}\right)=p \log_{b}(M)\). | \(=3 \log _{5} (4)\) |
b.
| \(\quad\log \left(x^{10}\right)\) | |
|---|---|
| Use the Power Property, \(\log_{b} \left(M^{p}\right)=p \log_{b} (M)\). | \(=10\log (x)\) |
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
a. \(\log _{7} \left(5^{4}\right)\)
b. \(\log \left(x^{100}\right)\)
- Answer
-
a. \(4\log _{7} (5)\)
b. \(100 \log (x)\)
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
a. \(\log _{2} \left(3^{7}\right)\)
b. \(\log \left(x^{20}\right)\)
- Answer
-
a. \(7\log _{2} (3)\)
b. \(20\log (x)\)
We summarize the Properties of Logarithms here for easy reference. While the natural logarithms are a special case of these properties, it is often helpful to also show the natural logarithm version of each property.
Properties of Logarithms
If \(M>0\), \(b>0\), \(b \neq 1\) and \(p\) is any real number, then
| Property | Base \(b\) | Base \(e\) |
|---|---|---|
| \(\log_{b} (1)=0\) | \(\ln (1)=0\) | |
| \(\log_{b} (b)=1\) | \(\ln (e)=1\) | |
| Inverse Properties |
\(b^{\log_{b} (x)}=x\)
\(\log_{b} \left(b^{x}\right)=x\) |
\(e^{\ln (x)}=x\)
\(\ln \left(e^{x}\right)=x\) |
| Product Property of Logarithms | \(\log_{b}(M N)=\log_{b} (M)+\log_{b} (N)\) | \(\ln (M N)=\ln (M)+\ln (N)\) |
| Quotient Property of Logarithms | \(\log_{b} \left(\dfrac{M}{N}\right)=\log_{b} (M)-\log_{b} (N)\) | \(\ln \left(\dfrac{M}{N}\right)=\ln (M)-\ln (N)\) |
| Power Property of Logarithms | \(\log_{b} \left(M^{p}\right)=p \log_{b} (M)\) | \(\ln \left(M^{p}\right)=p \ln (M)\) |
Now that we have the properties we can use them to “expand” a logarithmic expression. This means to write the logarithm as a sum or difference and without any powers.
We generally apply the Product and Quotient Properties before we apply the Power Property.
Use the Properties of Logarithms to expand the logarithm \(\log _{4}\left(2 x^{3} y^{2}\right)\). Simplify, if possible.
Solution
| \(\quad \log _{4}\left(2 x^{3} y^{2}\right)\) | |
|---|---|
| Use the Product Property, \(\log_{b} (M N)=\log_{b} (M)+\log_{b} (N)\). | \(=\log _{4}\left(2\right) +\log_4\left(x^{3}\right)+\log_4\left( y^{2}\right)\) |
| Use the Power Property, \(\log_{b} \left(M^{p}\right)=p \log_{b} (M)\), on the last two terms. |
\(=\log _{4}\left(4^{1/2}\right) +3\log_4\left(x\right)+2\log_4\left(y\right)\) \(=\dfrac{1}{2}\log _{4}(4) +3\log_4\left(x\right)+2\log_4\left(y\right)\) |
| Simplify. | \(=\dfrac{1}{2}+\dfrac{3}{2}\log_4\left(x\right)+\log_4\left(y\right)\) |
Use the Properties of Logarithms to expand the logarithm \(\log _{2}\left(5 x^{4} y^{2}\right)\). Simplify, if possible.
- Answer
-
\(\log _{2} (5)+4 \log _{2} (x)+2 \log _{2} (y)\)
Use the Properties of Logarithms to expand the logarithm \(\log _{3}\left(7 x^{5} y^{3}\right)\). Simplify, if possible.
- Answer
-
\(\log _{3} (7)+5 \log _{3} (x)+3 \log _{3} (y)\)
When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.
Use the Properties of Logarithms to expand the logarithm \(\log _{2} \left(\sqrt[4]{\dfrac{x^{3}}{3 y^{2} z}}\right)\). Simplify, if possible.
Solution
| \(\log _{2} \left(\sqrt[4]{\dfrac{x^{3}}{3 y^{2} z}}\right)\) | |
|---|---|
| Rewrite the radical with a rational exponent. | \(=\log _{2}\left(\dfrac{x^{3}}{3 y^{2} z}\right)^{1/4}\) |
|
Use the Power Property, \(\log_{b} \left(M^{p}\right)=p \log_{b} (M)\). |
\(=\dfrac{1}{4} \log _{2}\left(\dfrac{x^{3}}{3 y^{2} z}\right)\) |
| Use the Quotient Property, \(\log_{b} (M N)=\log_{b} (M)-\log_{b} (N)\). | \(=\dfrac{1}{4}\left(\log _{2}\left(x^{3}\right)-\log _{2}\left(3 y^{2} z\right)\right)\) |
| Use the Product Property, \(\log_{b}(MN)=\log_{b} (M)+\log_{b} (N)\), in the second term. |
\(=\dfrac{1}{4}\left(\log _{2}\left(x^{3}\right)-\left(\log _{2} (3)+\log _{2} \left(y^{2}\right)+\log _{2} (z)\right)\right)\) |
|
Use the Power Property, \(\log_{b} \left(M^{p}\right)=p \log_{b}(M)\), inside the parentheses. |
\(=\dfrac{1}{4}\left(3 \log _{2} (x)-\left(\log _{2} (3)+2 \log _{2} (y)+\log _{2} (z)\right)\right)\) |
| Simplify by distributing. | \(=\dfrac{3}{4} \log _{2} (x)-\dfrac{1}{4}\log _{2} (3)-\dfrac{1}{2} \log _{2} (y)-\dfrac{1}{4}\log _{2} (z)\) |
Use the Properties of Logarithms to expand the logarithm \(\log _{4} \left(\sqrt[5]{\dfrac{x^{4}}{2 y^{3} z^{2}}}\right)\). Simplify, if possible.
- Answer
-
\(\dfrac{4}{5} \log _{4} (x)-\dfrac{1}{10}-\dfrac{3}{5} \log _{4} (y)-\dfrac{2}{5} \log _{4} (z)\)
Use the Properties of Logarithms to expand the logarithm \(\log _{3} \left(\sqrt[3]{\dfrac{x^{2}}{5 y z}}\right)\). Simplify, if possible.
- Answer
-
\(\dfrac{2}{3} \log _{3} (x)-\dfrac{1}{3}\log _{3} (5)-\dfrac{1}{3}\log _{3} (y)-\dfrac{1}{3}\log _{3} (z)\)
The opposite of expanding a logarithm is to condense a sum or difference of logarithms that have the same base into a single logarithm. We again use the properties of logarithms to help us, but in reverse.
To condense logarithmic expressions with the same base into one logarithm, we start by using the Power Property to get the coefficients of the log terms to be one and then the Product and Quotient Properties as needed.
Use the Properties of Logarithms to condense the logarithm \(\log _{4} (3)+\log _{4} (x)-\log _{4} (y)\). Simplify, if possible.
Solution
| \(\quad\log _{4} (3)+\log _{4} (x)-\log _{4} (y)\) | |
|---|---|
| The log expressions all have the same base, \(4\). | |
| The first two terms are added, so we use the Product Property, \(\log_{b} (M)+\log_{b} (N)=\log_{b} (M N)\). | \(=\log _{4}(3x)-\log _{4} (y)\) |
| Since the logs are subtracted, we use the Quotient Property, \(\log_{b} (M)-\log_{b} (N)=\log_{b} \left(\dfrac{M}{N}\right)\). | \(=\log _{4}\left(\dfrac{3x}{y}\right)\) |
Use the Properties of Logarithms to condense the logarithm \(\log _{2} (5)+\log _{2} (x)-\log _{2} (y)\). Simplify, if possible.
- Answer
-
\(\log _{2} \left(\dfrac{5 x}{y}\right)\)
Use the Properties of Logarithms to condense the logarithm \(\log _{3} (6)-\log _{3}(x)-\log _{3} (y)\). Simplify, if possible.
- Answer
-
\(\log _{3} \left(\dfrac{6}{x y}\right)\)
Use the Properties of Logarithms to condense the logarithm \(2 \log _{3} (x)+4 \log _{3}(x+1)\). Simplify, if possible.
Solution
| \(\quad 2 \log _{3}(x)+4 \log _{3}(x+1)\) | |
|---|---|
| The log expressions have the same base, \(3\). | \(=2 \log _{3} (x)+4 \log _{3}(x+1)\) |
| Use the Power Property, \(\log_{b} (M)+\log_{b} (N)=\log_{b} (M N)\). | \(=\log _{3} \left(x^{2}\right)+\log _{3}\left((x+1)^{4}\right)\) |
| The terms are added, so we use the Product Property, \(\log_{b} (M)+\log_{b} (N)=\log_{b} (M N)\). | \(=\log _{3} \left(x^{2}(x+1)^{4}\right)\) |
Use the Properties of Logarithms to condense the logarithm \(3 \log _{2} (x)+2 \log _{2}(x-1)\). Simplify, if possible.
- Answer
-
\(\log _{2} \left(x^{3}(x-1)^{2}\right)\)
Use the Properties of Logarithms to condense the logarithm \(2 \log (x)+2 \log (x+1)\). Simplify, if possible.
- Answer
-
\(\log \left(x^{2}(x+1)^{2}\right)\)
Use the Change-of-Base Formula
To evaluate a logarithm with any other base, we can use the Change-of-Base Formula . We will show how this is derived.
| \(\quad\log_{a}(M)\) | |
|
Suppose we want to evaluate \(\log_{a}(M)\). Set it to be equal to \(y\). |
\(y=\log_{a}(M)\) |
| Rewrite the expression in exponential form. | \(a^{y}=M\) |
| Take the \(\log_{b}\) of each side. | \(\log_{b}\left(a^{y}\right)=\log_{b}(M)\) |
| Use the Power Property. | \(y\log_{b}(a)=\log_{b}(M)\) |
| Solve for \(y\). | \(y=\dfrac{\log_{b}(M)}{\log_{b}(a)}\) |
| Substitute}\(y=\log_{a}(M)\) and conclude. | \(\log_{a}(M)=\dfrac{\log_{b}(M)}{\log_{b}(a)}\) |
The Change-of-Base Formula introduces a new base \(b\). This can be any base \(b\) we want where \(b>0,b≠1\). Because our calculators have keys for logarithms base \(10\) and base \(e\), we will rewrite the Change-of-Base Formula with the new base as \(10\) or \(e\).
For any logarithmic bases \(a, b\) and \(M>0\),
\(\begin{array}{lll}{\log _{a} (M)=\dfrac{\log _{b} (M)}{\log _{b} (a)}} & {\log _{a} (M)=\dfrac{\log (M)}{\log (a)}} & {\log _{a} (M)=\dfrac{\ln (M)}{\ln (a)}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)
When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol \((≈)\).
Rounding to three decimal places, approximate \(\log _{4} (35)\).
Solution
| \(\quad\log _{4} (35)\) | |
|---|---|
| Use the Change-of-Base Formula, \(\log _{a} (M)=\dfrac{\log _{b} (M)}{\log _{b} (a)}\). | |
| Identify \(a\) and \(M\). Choose \(10\) for \(b\). | \(=\dfrac{\log(35)}{\log (4)}\) |
| Enter the expression \(\dfrac{\log (35)}{\log (4)}\) in the calculator using the log button for base \(10\). Round to three decimal places. |
\(\approx 2.565\) |
Rounding to three decimal places, approximate \(\log _{3} (42)\).
- Answer
-
\(3.402\)
Rounding to three decimal places, approximate \(\log _{5} (46)\).
- Answer
-
\(2.379\)
- In \(3^x=4\), does the base of the logarithm used when solve that equation matter?
- How do the properties of logarithms correspond to properties of exponents?
1. Assuming that \(x,y,z>0\), combine to an expression with one logarithm only.
\[\dfrac{2}{3}\log_5(x) - 5 \log_5(y)-3\log_5(z)\nonumber\]
2. Assuming that \(x,y>0\), expand
\[\log_6\left(\dfrac{\sqrt[5]{x^2}}{36y^3}\right).\nonumber\]
Key Concepts
- \(\log _{a} 1=0 \quad \log _{a} a=1\)
-
Inverse Properties of Logarithms
-
For \(a>0,x>0\) and \(a≠1\)
\(a^{\log _{a} x}=x \quad \log _{a} a^{x}=x\)
-
For \(a>0,x>0\) and \(a≠1\)
-
Product Property of Logarithms
-
If \(M>0,N>0,a>0\) and \(a≠1\), then,
\(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\)
The logarithm of a product is the sum of the logarithms.
-
If \(M>0,N>0,a>0\) and \(a≠1\), then,
-
Quotient Property of Logarithms
-
If \(M>0, N>0, \mathrm{a}>0\) and \(a≠1\), then,
\(\log _{a} \dfrac{M}{N}=\log _{a} M-\log _{a} N\)
The logarithm of a quotient is the difference of the logarithms.
-
If \(M>0, N>0, \mathrm{a}>0\) and \(a≠1\), then,
-
Power Property of Logarithms
-
If \(M>0,a>0,a≠1\) and \(p\) is any real number then,
\(\log _{a} M^{p}=p \log _{a} M\)
The log of a number raised to a power is the product of the power times the log of the number.
-
If \(M>0,a>0,a≠1\) and \(p\) is any real number then,
-
Properties of Logarithms Summary
If \(M>0,a>0,a≠1\) and \(p\) is any real number then,
| Property | Base \(a\) | Base \(e\) |
|---|---|---|
| \(\log _{a} 1=0\) | \(\ln 1=0\) | |
| \(\log _{a} a=1\) | \(\ln e=1\) | |
| Inverse Properties |
\(a^{\log _{a} x}=x\)
\(\log _{a} a^{x}=x\) |
\(e^{\ln x}=x\)
\(\ln e^{x}=x\) |
| Product Property of Logarithms | \(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\) | \(\ln (M \cdot N)=\ln M+\ln N\) |
| Quotient Property of Logarithms | \(\log _{a} \dfrac{M}{N}=\log _{a} M-\log _{a} N\) | \(\ln \dfrac{M}{N}=\ln M-\ln N\) |
| Power Property of Logarithms | \(\log _{a} M^{p}=p \log _{a} M\) | \(\ln M^{p}=p \ln M\) |
-
Change-of-Base Formula
For any logarithmic bases \(a\) and \(b\), and \(M>0\),\(\begin{array}{ll}{\log _{a} M=\dfrac{\log _{b} M}{\log _{b} a}} & {\log _{a} M=\dfrac{\log M}{\log a}} & {\log _{a} M=\dfrac{\ln M}{\ln a}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)