4.5: Logarithmic Properties
- Page ID
- 34906
Learning Objectives
- Use the product, quotient, and power rules for logarithms.
- Expand logarithmic expressions.
- Condense logarithmic expressions.
- Use the change-of-base formula for logarithms.
In chemistry, the pH scale is used as a measure of the acidity or alkalinity of a substance. Substances with a pH less than \(7\) are considered acidic, and substances with a pH greater than \(7\) are said to be alkaline. To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where \(a\) is the concentration of hydrogen ion in the solution
\[ {pH}=−{\log}([H^+]) ={\log}\left(\dfrac{1}{[H^+]}\right) \nonumber\]
The equivalence of these expressions is one of the logarithm properties we will examine in this section.
Basic Properties
Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here.
PROPERTIES OF ONE FOR LOGARITHMS
\( \log_b(1) =0 \qquad\qquad\qquad \log_b(b) =1 \qquad\qquad\qquad \ln(1) =0 \qquad\qquad\qquad \ln(e) =1 \)
These properties are easy to prove by writing the log equation in its exponential form.
\( \log_b(1) =0 \longleftrightarrow b^0 = (1) {\color{Cerulean}{✓}} \qquad\qquad\qquad \ln(1) =0 \longleftrightarrow \log_e(1) =0 \longleftrightarrow e^0 = (1) {\color{Cerulean}{✓}} \)
\( \log_b(b) =1 \longleftrightarrow b^1 = (b) {\color{Cerulean}{✓}} \qquad\qquad\qquad \ln(e) =1 \longleftrightarrow \log_e(e) =1 \longleftrightarrow e^1 = (e) {\color{Cerulean}{✓}} \)
Previously, a logarithm was evaluated by rewriting it in exponential form. But recognizing and applying log properties saves time!
Example \(\PageIndex{1}\)
Evaluate using the properties of logarithms:
- \(\log _{8} (1)\)
- \(\log _{6} (6)\)
Solution:
a. Use the property, \(\log _{b} (1)=0\): the log of 1, regardless of the base used is always zero \( \longrightarrow \log _{8} (1)=0 \)
\(\quad\) Previous approach: rewrite in exponential form: \(\log _{8} (1)=x \longleftrightarrow 8^x=1\longleftrightarrow x=0 \longleftrightarrow \log _{8} (1)=0 \)
b. Use the property, \(\log _{b} (b)=1\). Thus \(\log _{6} (6) = 1\)
Try It \(\PageIndex{1}\)
Evaluate using properties of logarithms:
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INVERSE PROPERTIES FOR LOGARITHMS
\( \log_b(b^x) =x \qquad\qquad\qquad b^{\log_b (x)}=x \qquad\qquad\qquad \ln(e^x) =x \qquad\qquad\qquad e^{\ln(x)} =x\)
The rational behind these properties is the fact that the functions \(f(x)=\log_b(x)\) and \(g(x)=b^x\) are inverse functions. As inverse functions \(f\) and \(g\) have the property that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\).
\((f \circ g)(x)=x \longleftrightarrow f(g(x))=x \longleftrightarrow f( b^x ) =x \longleftrightarrow {\color{Cerulean}{\log_b(b^x)=x \:✓}} \)
\((g \circ f)(x)=x \longleftrightarrow g(f(x))=x \longleftrightarrow g(\log_b(x)) =x \longleftrightarrow {\color{Cerulean}{b^{\log_b(x)} =x \:✓}} \)
Example \(\PageIndex{2}\)
Evaluate using properties of logarithms:
- \(4^{\log _{4} (9)}\)
- \(\log _{3} (3^{5})\)
Solution:
a. Use the property, \(b^{\log _{b} (x)}=x\). Therefore, \( \quad 4^{\log _{4} (9)}=9\)
b. Use the property, \(\log _{b} (b^x)=x\). Therefore, \( \quad \log _{3} (3^{5)}=5\)
Try It \(\PageIndex{2}\)
Evaluate using properties of logarithms:
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Using the Product Rule for Logarithms
Recall that we use the product rule of exponents to combine the product of exponents by adding: \(x^ax^b=x^{a+b}\). We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.
Given any real number \(x\) and positive real numbers \(M\), \(N\), and \(b\), where \(b≠1\), we will show
\({\log}_b(MN)={\log}_b(M)+{\log}_b(N)\).
Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b(MN)&= {\log}_b(b^mb^n) \qquad &&\text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m+n}) \qquad &&\text{Apply the product rule for exponents}\\[4pt] &= m+n \qquad &&\text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)+{\log}_b(N) \qquad &&\text{Substitute for m and n} \end{align*}\]
Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider \({\log}_b(wxyz)\). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
\({\log}_b(wxyz)={\log}_b(w)+{\log}_b(x)+{\log}_b(y)+{\log}_b(z)\)
PRODUCT RULE: The Log of a Product is the Sum of Logs
\[\begin{align*} {\log}_b(MN)={\log}_b(M)+{\log}_b(N)\text{ for } b> 0 \end{align*}\]
Example \(\PageIndex{3}\)
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible:
- \(\log _{3} (7 x)\)
- \(\log _{4} (64 x y)\)
- \({\log}_3(30x(3x+4))\)
Solution:
a. \(\quad\) Use the Product Property, \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\).
\( \begin{array}{rl}
\log _{3} (7 x) &= \log _{3} (7 \cdot x) \\
&= \log _{3} (7)+\log _{3} (x) \\
\end{array} \)
b. \(\quad\) Use the Product Property, \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\).
\( \begin{array}{rl}
\log _{4} (64 x y) &= \log _{4} (64 \cdot x \cdot y) \\
&=\log _{4} (64)+\log _{4} (x)+\log _{4} (y) \\
\end{array} \)
The argument \((64)\) is a power of the base, so it must be simplified. Rewrite and use the Inverse Property \( \log_b(b^x) =x \).
\( \begin{array}{rl}
\log _{4} (64 x y) &=\log _{4} (4^3)+\log _{4} (x)+\log _{4} (y) \\
&=3+\log _{4} (x)+\log _{4} (y) \\
\end{array} \)
c. \(\quad\)Use the Product Property, \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\).
\( \begin{array}{rl}
\log_3(30x(3x+4)) &=\log_3((30x)⋅(3x+4)) \\
&=\log_3(30x) + \log_3(3x+4) \\
\end{array} \)
The argument \((30x)\) is not a single factor, so it must be simplified. Use the Product Property again.
\( \begin{array}{rl}
\log_3(30x(3x+4)) &=\log_3(30 \cdot x) + \log_3(3x+4) \\
&=\log_3(30) + \log_3(x) + \log_3(3x+4) \\
\end{array} \)
However, the expression is still not completely simplified because the argument \((30)\) is a power of the base \((3)\). So write \(30\) as a product involving \(3\) as one of its factors. Then use the Product Property once more, followed by the property of One: \( \log_b(b) =1 \).
\( \begin{array}{rl}
\log_3(30x(3x+4)) &=\log_3(3 \cdot 10) + \log_3(x) + \log_3(3x+4) \\
&=\log_3(3) +\log_3(10) + \log_3(x) + \log_3(3x+4) \\
&=1 +\log_3(10) + \log_3(x) + \log_3(3x+4) \\
\end{array} \)
We will see in future that there are log properties for log arguments that are products, for log arguments that are quotients, and for log arguments that are powers. But, there is NO LOG PROPERTY for log arguments that are sums or differences. Therefore, \( \log_3(3x+4) \) CANNOT be simplified!!
Try It \(\PageIndex{3}\)
Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible:
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Using the Quotient Rule for Logarithms
For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: \(x^{\frac{a}{b}}=x^{a−b}\). The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms.
QUOTIENT RULE: The Log of a Quotient is the Difference of Logs
\({\log}_b\left(\dfrac{M}{N}\right)={\log}_bM−{\log}_bN\)
Just as with the product rule, we can use the inverse property to derive the quotient rule.
Given any real number \(x\) and positive real numbers \(M\), \(N\), and b, where \(b≠1\), we will show
\({\log}_b\left(\dfrac{M}{N}\right)={\log}_b(M)−{\log}_b(N)\).
Let \(m={\log}_bM\) and \(n={\log}_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
\[\begin{align*} {\log}_b\left (\dfrac{M}{N} \right )&= {\log}_b\left(\dfrac{b^m}{b^n}\right) \qquad &&\text{Substitute for M and N}\\[4pt] &= {\log}_b(b^{m-n}) \qquad &&\text{Apply the quotient rule for exponents}\\[4pt] &= m-n \qquad &&\text{Apply the inverse property of logs}\\[4pt] &= {\log}_b(M)-{\log}_b(N) \qquad &&\text{Substitute for m and n} \end{align*}\]
Example \(\PageIndex{4}\)
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
- \(\log _{5} (\frac{5}{7}) \)
- \(\log (\frac{x}{100})\)
Solution:
a. \(\quad\)Use the Quotient Property, \(\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N\).
\( \begin{array}{rll}
\log _{5} (\frac{5}{7}) &=\log _{5} (5)-\log _{5} (7) \\
&=1-\log _{5} 7 & \text{Simplify with Property of One: \( \log_b(b) =1\) }\\
\end{array} \)
b. \(\quad\)Use the Quotient Property, \(\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N\).
\( \begin{array}{rll}
\log (\frac{x}{100}) &=\log (x)-\log (100) \\
&=\log (x)-\log_{10}(10^2) & \text{ Use Inverse Property: \( \log_b(b^x) =x \) }\\
&=\log (x)-2 \\
\end{array} \)
Try It \(\PageIndex{4}\)
Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
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Using the Power Rule for Logarithms
We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as \(x^2\)? One method is as follows:
\[\begin{align*} {\log}_b(x^2)&= {\log}_b(x\cdot x)\\[4pt] &= {\log}_bx+{\log}_bx\\[4pt] &= 2{\log}_bx \end{align*}\]
Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power equals the product of the power and the log. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,
\(100={10}^2 \qquad\qquad\qquad \sqrt{3}=3^{\frac{1}{2}} \qquad\qquad\qquad \dfrac{1}{e}=e^{−1}\)
POWER RULE: The Log of a Power is the Product of the Power and the Log
\({\log}_b(M^p)=p{\log}_b(M)\)
Example \(\PageIndex{5}\)
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
- \(\log _{5} (4^{3})\)
- \(\log (x^{10})\)
- \({\log}_3(25)\)
Solution:
a. \(\quad\)Use the Power Property, \(\log _{b} M^{p}=p \log _{b} M \longrightarrow \log _{5} (4^{3})=3 \log _{5} 4\)
b. \(\quad\)Use the Power Property, \(\log _{b} M^{p}=p \log _{b} M \longrightarrow \log (x^{10})=10 \log (x) \).
c. \(\quad\)Use the Power Property, \(\log _{b} M^{p}=p \log _{b} M \longrightarrow {\log}_3(25)={\log}_3(5^2) = 2{\log}_3(5) \).
Try It \(\PageIndex{5}\)
Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
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e. \(\ln (x^2)\). f. \(\ln\left (\dfrac{1}{x^2} \right )\)
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Expanding Logarithmic Expressions
Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:
\[\begin{align*} {\log}_b \left (\dfrac{6x}{y} \right )&= {\log}_b(6x)-{\log}_by\\[4pt] &= {\log}_b6+{\log}_bx-{\log}_by \end{align*}\]
We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:
\[\begin{align*} {\log}_b\left (\dfrac{A}{C} \right )&= {\log}_b(AC^{-1})\\[4pt] &= {\log}_b(A)+{\log}_b(C^{-1})\\[4pt] &= {\log}_bA+(-1){\log}_bC\\[4pt] &= {\log}_bA−{\log}_bC \end{align*}\]
We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.
With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with arguments that are products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.
We generally apply the Product and Quotient Properties before we apply the Power Property.
Example \(\PageIndex{6}\): Logs of Products
Use the Properties of Logarithms to expand the following logarithms. Simplify, if possible.
- \(\log _{4}\left(2 x^{3} y^{2}\right)\)
- \( log_2(8x^4) \)
Solution:
a.
\( \begin{array}{rll}
\log _{4}\left(2 x^{3} y^{2}\right) &= \log _{4}(2) +\log _{4}(x^3) + \log _{4}(y^2) & \text{Product Property, \(\log _{b} M \cdot N=\log _{b} M+\log _{b} N\). } \\
&= \log _{4}(2) +3\log _{4}(x) + 2\log _{4}(y) & \text{Power Property, \({\log}_b(M^p)=p{\log}_b(M)\). } \\
\end{array} \)
b. In this problem notice that it is incorrect to use the power rule first! \( \log_2(8x^4) \ne 4 \log_2(8x) \)
\( \begin{array}{rll}
\log_2(8x^4) &= \log_2(8) + \log_2(x^4) & \text{Product Property, \(\log _{b} (M \cdot N)=\log _{b} M+\log _{b} N\). } \\
&= \log_2(2^3) + 4\log_2(x) & \text{Power Property, \({\log}_b(M^p)=p{\log}_b(M)\). } \\
&=3 + 4\log_2(x) & \text{Inverse Property, \( \log_b(b^x) =x \). }
\end{array} \)
Try It \(\PageIndex{6}\):
Use the Properties of Logarithms to expand the logarithms below. Simplify, if possible.
- \(\log _{2}\left(10 x^{4} y^{2}\right)\)
- \(\log _{3}\left(81 x^{5} y^{3}\right)\)
- Answer
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a. \(1+\log _{2} 5+4 \log _{2}(x)+2 \log _{2} (y) \)
b. \(4+5 \log _{3} x+3 \log _{3} y\)
Example \(\PageIndex{7}\): Logs of Quotients
Use the Properties of Logarithms to expand the following logarithms. Simplify, if possible.
- \( \ln \left (\dfrac{x^4y}{7} \right )\)
- \( \log_2 \Big( \dfrac{x^3}{4y} \Big) \)
- \({\log}_6 \left (\dfrac{64x^3(4x+1)}{(2x−1)} \right )\)
- \({\log}_2\left(\dfrac{15x(x−1)}{(3x+4)(2−x)}\right)\).
- \({\log}\left (\dfrac{2x^2+6x}{3x+9} \right ) \)
a. Solution
\( \begin{array}{rll}
\ln \left( \dfrac{x^4y}{7} \right) &=\ln(x^4y)−\ln(7) &\text{Quotient Rule \(\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N\) } \\
&=\ln(x^4 \cdot y)−\ln(7) \\
&=\ln(x^4)+\ln(y)−\ln(7) &\text{Product Rule \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\) } \\
&=4\ln(x)+\ln(y)−\ln(7) &\text{Power Rule \({\log}_b(M^p)=p{\log}_b(M)\) } \\
\end{array} \)
b. Solution
\( \begin{array}{rll}
\log_2 \Big( \dfrac{x^3}{4y} \Big) &=\log_2(x^3)-\log_2(4y) &\text{Quotient Rule \(\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N\) } \\
&=\log_2(x^3)-\log_2(4 \cdot y) \\
&=\log_2(x^3)- \big( \log_2(4) + \log_2(y) \big) &\text{Product Rule \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\) } \\
&=3\log_2(x)- 2 - \log_2(y) &\text{Power Rule \({\log}_b(M^p)=p{\log}_b(M)\); distribute; }\log_2(4)=2 \\
\end{array} \)
c. Solution
\( \begin{array}{rll}
{\log}_6 \left (\dfrac{64x^3(4x+1)}{(2x−1)} \right ) &={\log}_6(64)+{\log}_6(x^3)+{\log}_6(4x+1)-{\log}_6(2x-1) &\text{Product and Quotient Rules} \\
&={\log}_6(2^6)+{\log}_6(x^3)+{\log}_6(4x+1)-{\log}_6(2x-1) \\[4pt]
&=6{\log}_6(2)+3{\log}_6(x)+{\log}_6(4x+1)-{\log}_6(2x-1) &\text{Power Rule} \\
\end{array} \)
d. Solution
\( \begin{array}{rll}
{\log}_2\left(\dfrac{15x(x−1)}{(3x+4)(2−x)}\right) &={\log}_2(15x(x-1))-{\log}_2((3x+4)(2-x)) &\text{Quotient Rule } \\
&= [{\log}_2(15)+{\log}_2(x)+{\log}_2(x-1)]-[{\log}_2(3x+4)+{\log}_2(2-x)] &\text{Product Rule} \\[4pt]
&= {\log}_2(15)+{\log}_2(x)+{\log}_2(x-1)-{\log}_2(3x+4)-{\log}_2(2-x) &\text{Distribute} \\[4pt]
\end{array} \)
Analysis. There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for \(x=−\frac{4}{3}\) and \(x=2\). Also, since the argument of a logarithm must be positive, the expanded logarithm requires that \(x>0\), \(x>1\), \(x>−\frac{4}{3}\), and \(x<2\). Combining these conditions yields a domain of \(1<x<2\). In contrast the domain of the original log expression also includes the interval \( −\frac{4}{3}< x < 0 \). However, this analysis is beyond the scope of this section, and we will not consider these considerations here or in subsequent exercises.
e. Factoring and canceling we get,
\( \begin{array}{rll}
{\log}\left (\dfrac{2x^2+6x}{3x+9} \right ) &=\log(2x^2+6x ) -\log(3x+9 ) &\text{Quotient Rule} \\
&= \log(2x(x+3)) -\log(3(x+3)) &\text{Factor!!} \\
&= \log(2) + \log(x) + \log(x+3) -[\log(3) + \log(x+3)] &\text{Product Rule} \\
&= \log(2) + \log(x) + \log(x+3) -\log(3) - \log(x+3) &\text{Distribute negative} \\
&= \log(2) + \log(x) - \log(3) &\text{simplify} \\
\end{array} \)
Try It \(\PageIndex{7}\)
Use the Properties of Logarithms to expand the following logarithms. Simplify, if possible.
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a. \(\log \left (\dfrac{x^2y^3}{z^4} \right ) \\[4pt] \) b. \({\log}_3\left(\dfrac{7x^2+21x}{7x(x−1)(x−2)}\right) \\[4pt] \) c. \(\ln \left (\dfrac{\sqrt{(x−1){(2x+1)}^2}}{(x^2−9)}\right )\) |
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When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.
Example \(\PageIndex{8}\): Logs of Radicals
Use the Properties of Logarithms to expand the following logarithms. Simplify, if possible.
- \(\ln(\sqrt[3]{7x^2})\)
- \( \log _{2} \sqrt[4]{\dfrac{x^{3}}{3 y^{2} z}}\)
a. Solution
\( \begin{array}{rll}
\ln(\sqrt[3]{7x^2}) &= \ln (7x^2)^{\frac{1}{3}} &\text{Rewrite radical in exponential form} \\
&=\ln \Big( (7)^{\frac{1}{3}} \cdot (x^2)^{\frac{1}{3}} \Big) &\text{Use Power Rule for Exponents} \\
&=\ln (7)^{\frac{1}{3}} + \ln(x^ {\frac{2}{3}}) &\text{Product Rule \(\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N\)} \\
&=\frac{1}{3}\ln(7) + \frac{2}{3}\ln(x) &\text{Power Rule \({\log}_b(M^p)=p{\log}_b(M)\)} \\
\end{array} \)
b. Solution
\( \begin{array}{rll}
\log _2 \sqrt[4]{\frac{x^3}{3 y^2 z}}
&= \log _2 \left( \dfrac{x^3}{3 y^2 z} \right) ^{\tfrac{1}{4}}
&\text{Rewrite radical in exponential form} \\
&=\frac{1}{4} \log _2 \left( \dfrac{x^3}{3 y^2 z} \right)
&\text{Power Property, \( \log _b (M^p)=p \log _b (M)\). } \\
&=\frac{1}{4} \Big( \log _2 (x^3)-\log _2(3 y^2 z) \Big)
&\text{Quotient Property, \(\log _{b} \Big(\frac{M}{N}\Big)=\log _{b} (M)-\log _{b}(N)\) } \\
&= \frac{1}{4} \Big( 3 \log _2 (x) -\left( \log _2 (3) +2 \log _2 (y) +\log _2(z)\right) \Big)
&\text{Power & Product Property} \\
&=\frac{1}{4} \Big(3 \log _2 (x) -\log _2 (3) -2 \log _2 (y) -\log _{2}(z) \Big)
&\text{Simplify by distributing.} \\
&=\frac{3}{4}\log _2 (x) -\frac{1}{4}\log _2 (3) -\frac{1}{2} \log _2 (y) -\frac{1}{4}\log _2(z)
&\text{Write final result as a sum of individual terms.} \\
\end{array} \)
Try It \(\PageIndex{8}\)
Use the Properties of Logarithms to expand the following logarithms. Simplify, if possible.
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a. \(\log _{4} \sqrt[5]{\dfrac{x^{4}}{2 y^{3} z^{2}}}\).
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b. \(\log _{3} \sqrt[3]{\dfrac{x^{2}}{5 y z}}\).
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Can we expand \(\ln(x^2+y^2)\)?
No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Example \(\PageIndex{8x}\)
Given that \(\log _{2} x=a, \log _{2} y=b\), and that \(\log _{2} z=c\), write the following in terms of \(a\), \(b\), and \(c\):
- \(\log _{2}\left(8 x^{2} y\right)\)
- \(\log _{2}\left(\frac{2 x^{4}}{\sqrt{z}}\right)\)
Solution
- Begin by expanding using sums and coefficients and then replace \(a\) and \(b\) with the appropriate logarithm. \[\begin{aligned} \log _{2}\left(8 x^{2} y\right) &=\log _{2} 8+\log _{2} x^{2}+\log _{2} y \\ &=\log _{2} 8+2 \log _{2} x+\log _{2} y \\ &=3+2 a+b \end{aligned}\]
- Expand and then replace \(a, b\), and \(c\) where appropriate. \[\begin{aligned} \log _{2}\left(\frac{2 x^{4}}{\sqrt{z}}\right) &=\log _{2}\left(2 x^{4}\right)-\log _{2} z^{1 / 2} \\ &=\log _{2} 2+\log _{2} x^{4}-\log _{2} z^{1 / 2} \\ &=\log _{2} 2+4 \log _{2} x-\frac{1}{2} \log _{2} z \\ &=1+4 a-\frac{1}{2} b \end{aligned}\]
Condensing Logarithmic Expressions
We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.
How to: Condense a logarithmic expression into a single logarithm
Start with a sum and difference of logs all with the same base.
- Apply the reverse power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
- Next apply the reverse product property. Gather all the positive logs together and rewrite the sum of positive logarithms as the positive logarithm of a product. Gather all the negative logs together and rewrite the sum of negative logarithms as the negative logarithm of a product.
- Apply the reverse quotient property last. Rewrite the difference of logarithms as the logarithm of a quotient.
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.
>Example \(\PageIndex{9}\): Using the Log Properties in Reverse
Use the Properties of Logarithms to condense the logarithms below. Simplify, if possible.
a. \(4\ln(x)\)
b. \(2 \log _{3} x+4 \log _{3}(x+1)\)
c. \({\log}_3(5)+{\log}_3(8)−{\log}_3(2)\)
d. \(2\log x−4\log(x+5)+\dfrac{1}{x}\log(3x+5)\)
e. \( 5 \Big( \log_2(x^2)+\dfrac{1}{2}\log_2(x−1)−3\log_2 \Big( (x+3)^2 \Big) \Big) \)
f. \(\frac{1}{2} \ln x-3 \ln y-\ln z\)
g. \(-\ln(x) - \frac{1}{2}\)
a. Solution:
\(4\ln(x)=\ln(x^4) \qquad\qquad \text{ Reverse Power Rule \( p\log_b(M) = \log_b(M^p) \) } \)
b. Solution:
\( \begin{array}{rll}
2 \log _{3} x+4 \log _{3}(x+1) &=\log _{3} (x^2)+\log _{3}((x+1)^4) &\text{Reverse Power Rule \( p\log_b(M) = \log_b(M^p) \) } \\
&=\log _3(( x^2)(x+1)^4) &\text{Reverse Product Property, \( \log _b (M)+\log _b (N)=\log _b (M \cdot N) \). } \\
\end{array} \)
c. Solution
\( \begin{array}{rll}
{\log}_3(5)+{\log}_3(8)−{\log}_3(2) &= {\log}_3(5 \cdot 8)−{\log}_3(2) &\text{Reverse Product Rule } \\
&={\log}_3 \left (\dfrac{40}{2} \right ) &\text{Reverse Quotient Rule } \\
&={\log}_3(20) \\
\end{array} \)
d. Solution
\( \begin{array}{rll}
2\log x−4\log(x+5)+\dfrac{1}{x}\log(3x+5) &=\log(x^2)−\log{(x+5)}^4+\log({(3x+5)}^{x^{−1}}) &\text{Reverse power rule} \\
&=\log(x^2)+\log({(3x+5)}^{x^{-1}}-\log{(x+5)}^4 &\text{Rearrange } \\
&=\log(x^2{(3x+5)}^{x^{-1}})-\log{(x+5)}^4 &\text{Reverse Product Rule } \\
&=\log\dfrac{x^2{(3x+5)}^{1/x}}{{(x+5)}^4} &\text{Reverse Quotient Rule } \\
\end{array} \)
e. Solution
\( \begin{array}{rll}
5 \Big( \log_2(x^2)+\dfrac{1}{2}\log_2(x−1)−3\log_2(x+3)^2 \Big) &= 5{\log}_2(x^2)+\dfrac{5}{2}{\log}_2(x−1)−15{\log}_2(x+3)^2 &\text{Create a sum of terms} \\
&= {\log}_2(x^2)^5+{\log}_2(x−1)^{\tfrac{5}{2} }−{\log}_2({(x+3)}^2)^{15} &\text{Reverse Power rule} \\
&= \log_2(x^{10})+\log_2 (x−1)^{\tfrac{5}{2}} −\log_2(x+3)^{30} \\
&= \log_2(x^{10}(x−1)^{\frac{5}{2}} ) −\log_2(x+3)^{30} &\text{Reverse Product Rule} \\
&={\log}_2 \dfrac{x^2(x−1)^{\tfrac{5}{2}}}{(x+3)^{30}} &\text{Reverse Quotient Rule} \\
\end{array} \)
f. Solution
\( \begin{array}{rll}
\frac{1}{2} \ln x-3 \ln y-\ln z &=\ln x^{1 / 2} - \ln y^{3}-\ln z &\text{Reverse Power Rule } \\
&=\ln x^{1 / 2} - ( \ln y^{3} + \ln z) &\text{Combine negative terms} \\
&=\ln x^{1 / 2} - \ln (y^{3} z) &\text{Reverse Product Rule} \\
&=\ln \left(\dfrac{\sqrt{x}}{y^{3} z}\right) &\text{Reverse Quotient Rule } \\
\end{array} \)
g. Solution
\( \begin{array}{rcll}
-\ln(x) - \frac{1}{2} & = & (-1)\ln(x) - \frac{1}{2} & \\
& = & \ln\left(x^{-1}\right) - \frac{1}{2} & \mbox{Power Rule} \\
& = & \ln\left(x^{-1}\right) - \ln\left(e^{1/2}\right) & \mbox{Since \(\frac{1}{2} = \ln\left(e^{1/2}\right) \) } \\
& = & \ln\left(x^{-1}\right) - \ln\left(\sqrt{e} \right)& \\
& = & \ln\left(\dfrac{x^{-1}}{\sqrt{e}}\right) & \mbox{Quotient Rule} \\
& = & \ln\left(\dfrac{1}{x\sqrt{e}}\right) &
\end{array}\)
Try It \(\PageIndex{11a}\)
Use the Properties of Logarithms to condense the logarithms below. Simplify, if possible.
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a. \(2{\log}_34\) b. \(3 \log _{2} x+2 \log _{2}(x-1)\) c. \(2 \log x+2 \log (x+1)\) d. \(\log _{2} 5+\log _{2} x-\log _{2} y\) |
e. \(\log _{3} 6-\log _{3} x-\log _{3} y\) f. \({\log} 3−{\log} 4+{\log} 5−{\log} 6\) g. \(\log(5)+0.5\log(x)−\log(7x−1)+3\log(x−1)\) h. \(4(3\log(x)+\log(x+5)−\log(2x+3))\). |
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Using the Change-of-Base Formula for Logarithms
Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than \(10\) or, \(e\), we use the change-of-base formula to rewrite the logarithm as the quotient of common or natural logs.
To evaluate a logarithm with any other base, we can use the Change-of-Base Formula. We will show how this is derived.
\(\begin{array} {l c} {\text{Suppose we want to evaluate} \log_{a}M} & {\log_{a}M} \\ {\text{Let} \:y =\log_{a}M. }&{y=\log_{a}M} \\ {\text{Rewrite the epression in exponential form. }}&{a^{y}=M } \\ {\text{Take the }\:\log_{b} \text{of each side.}}&{\log_{b}a^{y}=\log_{b}M}\\ {\text{Use the Power Property.}}&{y\log_{b}a=\log_{b}M} \\ {\text{Solve for}\:y. }&{y=\dfrac{\log_{b}M}{\log_{b}a}} \\ {\text{Substitute}\:y=\log_{a}M.}&{\log_{a}M=\dfrac{\log_{b}M}{\log_{b}a}} \end{array}\)
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 \(e\), the base used with the Change-of-Base Formula when using a calculator is \(10\) or \(e\).
For example, to evaluate \({\log}_536\) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.
\[\begin{align*} {\log}_536&= \dfrac{\log(36)}{\log(5)} \qquad \text{Apply the change of base formula using base 10}\\[4pt] &\approx 2.2266 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}\]
THE CHANGE-OF-BASE FORMULA
The change-of-base formula can be used to evaluate a logarithm with any base.
For any positive real numbers \(M\), \(b\), and \(n\), where \(n≠1\) and \(b≠1\),
\({\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}\)
It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.
\[{\log}_bM=\dfrac{\ln M}{\ln b} \qquad\qquad\qquad \text{and} \qquad\qquad\qquad {\log}_bM=\dfrac{\log M}{\log b} \nonumber \]
How to: Rewrite \({\log}_bM\) as a quotient of logs with any positive base \(n\), where \(n≠1\)
- Determine the new base \(n\), remembering that the common log, \(\log(x)\), has base 10, and the natural log, \(\ln(x)\), has base \(e\).
- Rewrite the log as a quotient using the change-of-base formula
- The numerator of the quotient will be a logarithm with base \(n\) and argument \(M\).
- The denominator of the quotient will be a logarithm with base \(n\) and argument \(b\).
Example \(\PageIndex{12}\): Changing Logarithmic Expressions to Expressions Involving Only Natural Logs
Change \({\log}_53\) to a quotient of natural logarithms.
Solution
Because we will be expressing \({\log}_53\) as a quotient of natural logarithms, the new base, \(n=e\).
We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument \(3\). The denominator of the quotient will be the natural log with argument 5.
\({\log}_bM=\dfrac{\ln M}{\ln b}\)
\({\log}_53=\dfrac{\ln3}{\ln5}\)
Try It \(\PageIndex{12}\)
Change \(\log 0.58\) to a quotient of natural logarithms.
- Answer
-
\(\dfrac{\ln (.58)}{\ln (10)}\)
Can we change common logarithms to natural logarithms?
Yes. Remember that \(\log9\) means \({\log}_{10}9\). So, \(\log9=\dfrac{\ln9}{\ln10}\).
Example \(\PageIndex{13}\): Using the Change-of-Base Formula with a Calculator
Evaluate \({\log}_2(10)\) using the change-of-base formula with a calculator.
Solution
According to the change-of-base formula, we can rewrite the log base \(2\) as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base \(e\).
\[\begin{align*} {\log}_210&= \dfrac{\ln10}{\ln2} \qquad \text{Apply the change of base formula using base } e\\[4pt] &\approx 3.3219 \qquad \text{Use a calculator to evaluate to 4 decimal places} \end{align*}\]
Try It \(\PageIndex{13}\)
Evaluate \({\log}_5(100)\) using the change-of-base formula.
- Answer
-
\(\dfrac{\ln100}{\ln5}≈\dfrac{4.6051}{1.6094}=2.861\)
Applications of Logs
Example \(\PageIndex{14}\): Application of the Laws of Logs
Recall that, in chemistry, \({pH}=−\log[H+]\). If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?
Solution
Suppose \(C\) is the original concentration of hydrogen ions, and \(P\) is the original pH of the liquid. Then \(P= \: – \log(C)\). If the concentration is doubled, the new concentration is \(2C\). Then the pH of the new liquid is
\(pH=−\log(2C)\)
Using the product rule of logs
\(pH=−\log(2C)=−(\log(2)+\log(C))=−\log(2)−\log(C)\)
Since \(P=\:–\log(C)\), the new pH is
\(pH=P−\log(2)≈P−0.301\)
Try It \(\PageIndex{14}\)
When the concentration of hydrogen ions is doubled, the pH decreases by about \(0.301\).
How does the pH change when the concentration of positive hydrogen ions is decreased by half?
- Answer
-
The pH increases by about \(0.301\).
Key Equations
| The Product Rule for Logarithms | \({\log}_b(MN)={\log}_b(M)+{\log}_b(N)\) |
| The Quotient Rule for Logarithms | \({\log}_b(\dfrac{M}{N})={\log}_bM−{\log}_bN\) |
| The Power Rule for Logarithms | \({\log}_b(M^n)=n{\log}_bM\) |
| The Change-of-Base Formula | \({\log}_bM=\dfrac{{\log}_nM}{{\log}_nb}\) \(n>0\), \(n≠1\), \(b≠1\) |
Key Concepts
- We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example \(\PageIndex{1}\).
- We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example \(\PageIndex{2}\).
- We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example \(\PageIndex{3}\), Example \(\PageIndex{4}\), and Example \(\PageIndex{5}\).
- We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example \(\PageIndex{6}\), Example \(\PageIndex{7}\), and Example \(\PageIndex{8}\).
- The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example \(\PageIndex{9}\), Example \(\PageIndex{10}\), Example \(\PageIndex{11}\), and Example \(\PageIndex{12}\).
- We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See Example \(\PageIndex{13}\).
- The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and \(e\) as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example \(\PageIndex{14}\).
Contributors
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.