13.5: Laws of Logarithms

Learning Objectives Use the Product Law for logarithms. Use the Quotient Law for logarithms. Use the Power Law for logarithms. Expand logarithmic expressions. Condense logarithmic expressions. Use the Change-of-Base Formula for logarithms.

Hawk A.I. Section-Specific Tutor Please note that, to access the Hawk A.I. Tutor, you will need a (free) OpenAI account.

Figure \( \PageIndex{ 1 } \)

The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)
 

In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from \( 0 \) to \( 14 \). Substances with a pH less than \( 7 \) are considered acidic, and substances with a pH greater than \( 7 \) are said to be alkaline. Our bodies, for instance, must maintain a pH close to \( 7.35 \) in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:

• Battery acid: 0.8
• Stomach acid: 2.7
• Orange juice: 3.3
• Pure water: 7 (at \(25^{\circ}\) C)
• Human blood: 7.35
• Fresh coconut: 7.8
• Sodium hydroxide (lye): 14

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 \(H^+\), is the concentration of hydrogen ion in the solution\[ \text{pH} = -\log([H^+]) = \log\left( \dfrac{1}{[H^+]} \right). \nonumber \]The equivalence of \(-\log([H^+])\) and \( \log \left( \frac{1}{[H^+]}\right)\), is one of the Laws of Logarithms we will examine in this section.

Using the Product Law for Logarithms

Recall that the logarithmic and exponential functions "undo" each other. This means that logarithms have similar laws to exponents. Some important laws of logarithms are given here. First, the following "Base Laws" are easy to prove.

Theorem: Laws of Logarithms - Base Laws

\[ \log_b (1) = 0 \quad \text{and} \quad \log_b (b) = 1. \nonumber \]

Proofs

By definition of a logarithm, \( \log_b (1) \) is the exponent, \( p \), for which \( b^p = 1 \). Therefore, \( p = 0 \). As such,\[ \log_b(1) = 0. \nonumber \]Similarly, \( \log_b(b) \) is the exponent, \( q \), such that \( b^q = b \). Since \( b = b^1 \), this means that \( b^q = b^1 \). In other words, \( q = 1 \). Hence,\[ \log_b(b) = 1. \nonumber \]

For example, \( \log_5 1 = 0\), since \( 5^0 = 1 \), and \( \log_5 5 = 1\), since \( 5^1 = 5 \).

Next, we have the Inverse Laws.

Theorem: Laws of Logarithms - Inverse Laws

\[ \log_b ( b^x ) = x \quad \text{and} \quad b^{\log_b (x)} = x, \text{ where }x > 0. \nonumber \]

Proofs

Consider \( \log_b(b^x) \). By the definition of a logarithm, this is the exponent, \( p \), such that \( b^p = b^x \). That is, \( x = p \). Thus, \( \log_b(b^x) = x \).

Now consider \( b^{\log_b(x)} \), where \( x > 0 \). Recall that \( \log_b(x) \) is the exponent, \( q \), such that \( b^q = x \). That is,\[ b^{\log_b(x)} = b^q = x. \nonumber \]

To illustrate these laws, we imagine evaluating \( \log(100)\). We can rewrite the logarithm as \( \log_{10}( 10^2)\), and then apply the Inverse Law \( \log_b ( b^x ) = x\), to get \( \log_{10}(10^2) = 2\).

To evaluate \( e^{\ln( 7 )}\), we can rewrite the logarithm as \(e^{\log_e 7}\), and then apply the Inverse Law \(b^{\log_b x} =x\) to get \(e^{\log_e 7} = 7\).

Finally, we have the One-to-One Law.

Theorem: Laws of Logarithms - One-to-One Law

\[ \log_b (M) = \log_b (N) \text{ if and only if } M = N. \nonumber \]

Proof

From the definition of a logarithm, if we let \( \log_b(M) = m  \) and \( \log_b(N) =n \), then \( b^m = M \) and \( b^n = N \).

(\( \implies \)): If \( \log_b(M) = \log_b(N) \), then \( m = n \). However, if \( m = n \), then \( b^m = b^n \). Therefore, \( M = N \).

(\( \impliedby \)): If \( M = N \), then \( b^m = b^n \). However, by laws of exponents, this means \( m = n \). Hence, \( \log_b (M) = \log_b(N) \).

We can use the One-to-One Law to solve the equation \( \log_3 (3x) = \log_3 (2x+5)\), for \(x\). Since the bases are the same, we can apply the One-to-One Law by setting the arguments equal and solving for \(x\):\[ \begin{array}{rrclcl} & 3x & = & 2x + 5 & \quad & \left( \text{setting the arguments equal to each other} \right) \\[6pt] \implies & x & = & 5 & \quad & \left( \text{subtracting }2x\text{ from both sides} \right) \\[6pt] \end{array} \nonumber \]But what about the equation \( \log_3 (3x) + \log_3 (2x+5) = 2\)? The One-to-One Law does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.

Recall that we use the Product Law of Exponents to combine the product of powers by adding exponents: \(x^a x^b = x^{a + b} \). We have a similar law for logarithms, called the Product Law for Logarithms.

Theorem: Laws of Logarithms - Product Law

Given any real number \(x\), and positive real numbers \(M\) and \(N\),\[ \log_b(MN) = \log_b (M) + \log_b (N). \nonumber \]

Proof

Let \(m = \log_b (M)\), and \(n = \log_b (N)\). In exponential form, these equations are \(b^m = M\) and \(b^n = N\). It follows that\[ \begin{array}{rclcl} \log_b(MN) & = & \log_b(b^m b^n) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b(b^{m + n}) & \quad & \left( \text{Laws of Exponents: Product Rule} \right) \\[6pt] & = & m + n & \quad & \left( \text{Laws of Logarithms: Inverse Law} \right) \\[6pt] & = & \log_b(M) + \log_b(N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]

Note that repeated applications of the Product Law 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 Law for Logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:\[ \log_b (wxyz) = \log_b (w) + \log_b (x) + \log_b (y) + \log_b (z). \nonumber \]

Example \( \PageIndex{ 1 } \): Using the Product Law for Logarithms

Expand \( \log_3 ( 30x ( 3x+4 ) )\).

Solution

We begin by recognizing the argument as a product of three factors.\[ \log_3 ( 30x( 3x+4 ) )= \log_3 ( 30 \cdot x \cdot ( 3x+4 ) ). \nonumber \]Next we write the equivalent equation by summing the logarithms of each factor.\[ \log_3 ( 30 \cdot x \cdot ( 3x+4 ) )= \log_3 ( 30 ) + \log_3 ( x )+ \log_3 ( 3x+4 ). \nonumber \]Finally, we need to be aware of the base and the possibiliy that the \( 30 \) can be factored into a power of the base and some other constant. Specifically, \( 30 = 3 \cdot 10 \). Therefore,\[ \begin{array}{rclcl} \log_3 ( 30x( 3x+4 ) ) & = & \log_3 ( 3 \cdot 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & & \\[6pt] & = & \log_3 ( 3 ) + \log_3 ( 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & \quad & \left( \text{Laws of Logorithms: Product Law} \right) \\[6pt] & = & 1 + \log_3 ( 10 ) + \log_3 ( x )+ \log_3 ( 3x+4 ) & \quad & \left( \text{Laws of Logorithms: }\log_b b = 1 \right) \\[6pt] \end{array} \nonumber \]

Checkpoint \( \PageIndex{ 1 } \)

Expand \( \log_b (8k)\).

Using the Quotient Law for Logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the Quotient Law of Exponents to combine the quotient of exponents by subtracting: \( \frac{x^a}{x^b} = x^{a−b} \). The Quotient Law for Logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the Product Law, we can use the Inverse Law to derive the Quotient Law.

Theorem: Laws of Logarithms - Quotient Law

Given any real number \(x\), and positive real numbers \(M\) and \(N\),\[ \log_b \left( \dfrac{M}{N} \right) = \log_b ( M )− \log_b ( N ). \nonumber \]

Proof

Let \(m= \log_b (M)\), and \(n= \log_b (N)\). In exponential form, these equations are \( b^m = M \), and \(b^n = N\). It follows that\[ \begin{array}{rclcl} \log_b \left( \dfrac{M}{N} \right) & = & \log_b \left( \dfrac{b^m}{b^n} \right) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b \left( b^{m - n} \right) & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt] & = & m - n & \quad & \left( \text{Laws of Exponents: Inverse Law} \right) \\[6pt] & = & \log_b (M) - \log_b (N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]

For example, to expand \( \log \left( \frac{2x^2 + 6x}{3x + 9} \right)\), we must first express the quotient in lowest terms. Factoring and canceling we get,\[ \begin{array}{rclcl} \log \left( \dfrac{2x^2 + 6x}{3x + 9} \right) & = & \log \left( \dfrac{2x(x + 3)}{3(x + 3)} \right) & \quad & \left( \text{factoring out the GCF from numerator and denominator} \right) \\[6pt] & = & \log \left( \dfrac{2x \cancel{(x + 3)}}{3\cancel{(x + 3)}} \right) & \quad & \left( \text{canceling like factors} \right) \\[6pt] & = & \log \left( \dfrac{2x}{3} \right) & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]Next we apply the Quotient Law by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the Product Law.\[ \log\left( \dfrac{2x}{3} \right) = \log(2x) - \log(3) = \log(2) + \log(x) - \log(3). \nonumber \]

Caution: Assumptions Abound

When using the Product and Quotient Laws of Logarithms, we are assuming (often falsely) that the component logarithms are well-defined. That is, when expanding \( \log \left( \frac{2x^2 + 6x}{3x + 9} \right)\) to arrive at \( \log(2) + \log(x) - \log(3) \), we are assuming

1. \( x > 0 \), and
2. \( x + 3 \neq 0 \) (remember, this is the factor we canceled in the process of simplifying the logarithm).

These assumptions are likely not correct! However, when practicing the Laws of Logarithms, we allow ourselves the "training wheels" of assuming everything is going to work out. When we start to solve equations involving logarithms, on the other hand, we remove those "training wheels" and become concerned with the possibility that the component logarithms are not well-defined.

Example \( \PageIndex{ 2 } \): Using the Quotient Law for Logarithms

Expand\[ \log_2 \left( \dfrac{15x(x - 1)}{(3x + 4)(2 - x)} \right). \nonumber \]

Solution

First we note that the quotient is factored and in lowest terms, so we apply the Quotient Law.\[ \log_2 \left( \dfrac{15x(x - 1)}{(3x + 4)(2 - x)} \right) = \log_2 ( 15x(x−1) ) − \log_2 ( (3x+4)(2−x) ). \nonumber \]Notice that the resulting terms are logarithms of products. To expand completely, we apply the Product Law, noting that \( 15 \) is not divisible by a power of the base (\( 2 \)).\[ \begin{array}{rclcl} \log_2 ( 15x(x−1) ) − \log_2 ( (3x+4)(2−x) ) & = & \log_2 (15) + \log_2 (x) + \log_2(x−1) − \left[ \log_2(3x+4) + \log_2(2−x) \right] & \quad & \left( \text{Product Law} \right) \\[6pt] & = & \log_2 (15) + \log_2 (x) + \log_2(x−1) − \log_2(3x+4) - \log_2(2−x) & \quad & \left( \text{distributing} \right) \\[6pt] \end{array} \nonumber \]

Let's restate our "Caution" from above. There are exceptions to consider in Example \( \PageIndex{ 2 } \). First, because denominators must never be zero, the expression in Example \( \PageIndex{ 2 } \) is not defined for \(x= −\frac{4}{3}\) and \(x=2\). Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that \(x > 0\), \(x >1 \), \(x > −\frac{4}{3} \), and \(x < 2\). Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

Checkpoint \( \PageIndex{ 2 } \)

Expand \( \log_3 \left( \frac{7x^2 + 21x}{7x(x - 1)(x - 2)} \right)\).

Using the Power Law for Logarithms

We’ve explored the Product Law and the Quotient Law, but how can we take the logarithm of a power, such as \( x^2 \)? One method is as follows:\[ \begin{array}{rclcl} \log_b(x^2) & = & \log_b(x \cdot x) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] & = & \log_b x + \log_b x & \quad & \left( \text{Laws of Logarithms: Product Law} \right) \\[6pt] & = & 2 \log_b x & \quad & \left( \text{combining like terms} \right) \\[6pt] \end{array} \nonumber \]Notice that we used the Product Law for Logarithms to find a solution for the example above. By doing so, we have derived the Power Law for Logarithms, which says that the log of a power is equal to the exponent times the log of the base. 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,\[100 = 10^2 \quad \sqrt{3} = 3^{1/2} \quad \dfrac{1}{e} = e^{−1}. \nonumber \]

Laws of Logarithms: Power Law

\[ \log_b ( M^p ) = p \log_b (M). \nonumber \]

Proof

Let \( \log_b (M) = m \).\[ \begin{array}{rrclcl}
\implies & b^m & = & M & \quad & \left( \text{definition of the logarithm} \right) \\[6pt]
\implies & \left( b^m \right)^p & = & M^p & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & b^{mp} & = & M^p & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \log_b \left(b^{mp}\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{One-to-One Law of Logarithms} \right) \\[6pt]
\implies & \log_b \left(b^m \cdot b^p\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \log_b \left(b^m \right) \cdot \log_b \left( b^p\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{Product Law of Logarithms} \right) \\[6pt]
\implies & m \cdot p & = & \log_b \left(M^p\right) & \quad & \left( \text{Base Laws of Logarithms} \right) \\[6pt]
\implies & \log_b(M) \cdot p & = & \log_b \left(M^p\right) & \quad & \left( \text{our original substitution }\log_b(M) = m \right) \\[6pt]
\implies & p \log_b(M) & = & \log_b \left(M^p\right) & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt]
\end{array} \nonumber \]

Example \( \PageIndex{ 3 } \): Expanding a Logarithm with Powers

Expand \( \log_2 (x^5) \).

Solution

The argument is already written as a power, so we identify the exponent, \( 5 \), and the base, \(x\), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\[ \log_2 ( x^5 ) =5 \log_2 x \nonumber \]

Checkpoint \( \PageIndex{ 3 } \)

Expand \(\ln (x^2) \).

Example \( \PageIndex{ 4 } \): Rewriting an Expression as a Power before Using the Power Law

Expand \( \log_3 ( 25 )\), using the Power Law for Logs.

Solution

Expressing the argument as a power, we get \( \log_3 ( 25 )= \log_3 ( 5^2 )\).

Next we identify the exponent, \( 2 \), and the base, \( 5 \), and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.\[ \log_3 ( 5^2 ) = 2 \log_3 ( 5 ) \nonumber \]

Checkpoint \( \PageIndex{ 4 } \)

Expand \(\ln\left( \frac{1}{x^2} \right)\).

Example \( \PageIndex{ 5 } \): Using the Power Law in Reverse

Rewrite \(4\ln(x)\), using the Power Law for logs to a single logarithm with a leading coefficient of \( 1 \).

Solution

Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression \(4\ln(x)\), we identify the factor, \( 4 \), as the exponent and the argument, \(x\), as the base, and rewrite the product as a logarithm of a power: \(4 \ln(x)= \ln( x^4 )\).

Checkpoint \( \PageIndex{ 5 } \)

Rewrite \(2 \log_3 4\), using the Power Law for logs to a single logarithm with a leading coefficient of 1.

Expanding Logarithmic Expressions

Taken together, the laws introduced in this section so far are collectively called the "Laws of Logarithms." Sometimes we apply more than one rule in order to simplify an expression. For example:\[ \begin{array}{rcl} \log_b\left( \dfrac{6x}{y} \right) & = & \log_b(6x) - \log_b y \\[6pt] & = & \log_b 6 + \log_b x - \log_b y \\[6pt] \end{array} \nonumber \]We can use the Power Law to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the Quotient Law for Logarithms using the fact that a reciprocal is a negative power:\[ \begin{array}{rcl} \log_b\left( \dfrac{A}{C} \right) & = & \log_b(A C^{-1}) \\[6pt] & = & \log_b (A) + \log_b (C^{-1}) \\[6pt] & = & \log_b A + (-1) \log_b C \\[6pt] & = & \log_b A - \log_b C \\[6pt] \end{array} \nonumber \]We can also apply the Product Law 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 products, quotients, powers, and roots — never with addition or subtraction inside the argument of the logarithm.

Example \( \PageIndex{ 6 } \): Expanding Logarithms Using Product, Quotient, and Power Laws

Rewrite \(\ln\left( \frac{x^4 y}{7} \right)\), as a sum or difference of logs.

Solution

First, because we have a quotient of two expressions, we can use the Quotient Law:\[\ln\left( \dfrac{x^4 y}{7} \right)= \ln( x^4 y ) − \ln(7). \nonumber \]Then seeing the product in the first term, we use the Product Law:\[ \ln( x^4 y ) − \ln(7) = \ln( x^4 ) + \ln(y) − \ln(7). \nonumber \]Finally, we use the Power Law on the first term:\[\ln( x^4 ) + \ln(y) − \ln(7) = 4 \ln(x) + \ln(y) − \ln(7). \nonumber \]

Checkpoint \( \PageIndex{ 6 } \)

Expand \( \log\left( \frac{x^2 y^3}{z^4} \right)\).

Example \( \PageIndex{ 7 } \): Using the Power Law for Logarithms to Simplify the Logarithm of a Radical Expression

Expand \( \log( \sqrt{ x } )\).

Solution

\[ \log( \sqrt{ x } ) = \log x^{1/2} = \dfrac{1}{2} \log x \nonumber \]

Checkpoint \( \PageIndex{ 7 } \)

Expand \(\ln( \sqrt[3]{x^2} )\).

Q&A

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 \( \PageIndex{ 8 } \): Expanding Complex Logarithmic Expressions

Expand \( \log_6 \left( \frac{64x^3 (4x + 1)}{2x - 1} \right)\).

Solution

We can expand by applying the Product and Quotient Laws.\[ \begin{array}{rcl} \log_6 \left( \frac{64x^3 (4x + 1)}{2x - 1} \right) & = & \log_6 64 + \log_6 x^3 + \log_6 (4x + 1) - \log_6 (2x - 1) \\[6pt] & = & \log_6 64 + 3\log_6 x + \log_6 (4x + 1) - \log_6 (2x - 1) \\[6pt] \end{array} \nonumber \]Note that we could have rewritten \( \log_6 64 \) as \( \log_6 2^6 = 6 \log_6 2 \), however, this is not any better than \( \log_6 64 \). When it comes to logarithms of constants, we are only concerned with the possibility that a power of the base is a factor of the constant.

Checkpoint \( \PageIndex{ 8 } \)

Expand \( \ln\left( \frac{\sqrt{(x-1)(2x + 1)^2}}{x^2 - 9} \right)\).

Condensing Logarithmic Expressions

We can use the Laws of Logarithms 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. Because students sometimes make mistakes on the order with which they apply the Laws of Logarithms, we provide the following "How To" as a beginner's reference guide.

How To

Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.

1. Apply the Power Law first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
2. Next apply the Product Law. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the Quotient Law last. Rewrite differences of logarithms as the logarithm of a quotient.

Example \( \PageIndex{ 9 } \): Using the Product and Quotient Laws to Combine Logarithms

Write \( \log_3 ( 5 )+ \log_3 ( 8 )− \log_3 ( 2 )\) as a single logarithm.

Solution

Using the product and Quotient Laws\[ \log_3 ( 5 )+ \log_3 ( 8 )= \log_3 ( 5 \cdot 8 )= \log_3 ( 40 ). \nonumber \]This reduces our original expression to\[ \log_3 (40)− \log_3 (2). \nonumber \]Then, using the Quotient Law\[ \log_3 ( 40 )− \log_3 ( 2 )= \log_3 \left( \dfrac{40}{2} \right)= \log_3 ( 20 ). \nonumber \]

Checkpoint \( \PageIndex{ 9 } \)

Condense \( \log 3 - \log 4 + \log 5 - \log 6\).

Example \( \PageIndex{ 10 } \): Condensing Complex Logarithmic Expressions

Condense \( \log_2 ( x^2 )+ \frac{1}{2} \log_2 ( x−1 )−3 \log_2 ( ( x+3 )^2 )\).

Solution

We apply the Power Law first:\[ \log_2 ( x^2 )+ \frac{1}{2} \log_2 ( x−1 )−3 \log_2 ( ( x+3 )^2 )= \log_2 ( x^2 )+ \log_2 ( \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 ). \nonumber \]Next we apply the Product Law to the sum:\[ \log_2 ( x^2 )+ \log_2 ( \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 )= \log_2 ( x^2 \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 ). \nonumber \]Finally, we apply the Quotient Law to the difference:\[ \log_2 ( x^2 \sqrt{ x−1 } )− \log_2 ( ( x+3 )^6 )= \log_2 \left( \dfrac{x^2 \sqrt{ x−1 }}{( x+3 )^6} \right). \nonumber \]

Checkpoint \( \PageIndex{ 10 } \)

Rewrite \( \log( 5 ) + 0.5 \log( x ) - \log( 7x−1 ) + 3\log( x−1 )\) as a single logarithm.

Example \( \PageIndex{ 11 } \): Rewriting as a Single Logarithm

Rewrite \(2 \log x − 4 \log(x+5) + \frac{1}{x} \log( 3x+5 )\) as a single logarithm.

Solution

We apply the Power Law first:\[2 \log x – 4 \log(x+5) + \dfrac{1}{x} \log(3x+5) = \log( x^2 ) - \log ( x+5 )^4 + log\left( (3x+5)^{x^{−1}} \right). \nonumber \]Next, we rearrange and apply the Product Law to the sum:\[ \begin{array}{rcl} \log( x^2 ) - \log ( x+5)^4 + \log\left( (3x+5)^{x^{−1}} \right) & = & \log( x^2 ) + \log\left( (3x+5)^{x^{−1}} \right) - \log ( x+5)^4 \\[6pt] & = & \log\left(x^2 (3x+5)^{x^{−1}} \right) - \log(x+5)^4 \\[6pt] \end{array} \nonumber \]Finally, we apply the Quotient Law to the difference:\[ = \log \left( x^2 ( 3 x + 5 )^{x^{−1}} \right) − \log ( x + 5 )^4 = \log \dfrac{x^2 ( 3 x + 5 )^{x^{−1}}}{( x + 5 )^4}. \nonumber \]

Checkpoint \( \PageIndex{ 11 } \)

Condense \(4( 3 \log( x ) + \log( x+5 ) - \log( 2x+3 ) )\).

Example \( \PageIndex{ 12 } \): Applying of the Laws of Logs

Recall that, in chemistry, \(\text{pH} = -\log[ H^+ ]\). 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\[\text{pH}=-\log( 2C ). \nonumber \]Using the Product Law of logs\[\text{pH} = -\log( 2C ) = −( \log(2) + \log(C) ) = -\log(2) - \log(C). \nonumber \]Since \(P = –\log(C)\), the new pH is\[\text{pH} = P - \log(2) \approx P − 0.301. \nonumber \]When the concentration of hydrogen ions is doubled, the pH decreases by about \( 0.301 \).

Checkpoint \( \PageIndex{ 12 } \)

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

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 logarithms of any other base; when using a calculator, we would change them to common or natural logs.

Theorem: Change-of-Base Formula

Given any positive real numbers \(M\), \(b\), and \(n\), where \(n \neq 1\), and \(b \neq 1\),\[ \log_b (M) = \dfrac{\log_n (M)}{\log_n (b)}. \nonumber \]

Proof

Let \(y= \log_b (M)\). By the definition of a logarithm, \(b^y =M\). It follows that\[ \begin{array}{rrclcl}
& \log_n ( b^y ) & = & \log_n (M) & \quad & \left( \text{One-to-One Law of Logarithms} \right) \\[6pt]
\implies & y \log_n (b) & = & \log_n (M) & \quad & \left( \text{Power Law of Logarithms} \right) \\[6pt]
\implies & y & = & \dfrac{\log_n (M)}{\log_n (b)} & \quad & \left( \text{dividing both sides by }\log_n (b) \right) \\[6pt]
\implies & \log_b (M) & = & \dfrac{\log_n (M)}{\log_n (b)} & \quad & \left( \text{substituting} \right) \\[6pt]
\end{array} \nonumber \]

For example, to evaluate \( \log_5 36\) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.\[ \log_5 36 = \dfrac{\log( 36 )}{\log( 5 )} \approx 2.2266. \nonumber \]

Example \( \PageIndex{ 13 } \): Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Change \( \log_5 (3)\) to a quotient of natural logarithms.

Solution

Because we will be expressing \( \log_5 3\) 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 \( 5 \).\[ \log_5 3 = \dfrac{\ln 3}{\ln 5}. \nonumber \]

Checkpoint \( \PageIndex{ 13 } \)

Change \( \log_{0.5} (8)\) to a quotient of natural logarithms.

Q&A

Can we change common logarithms to natural logarithms?

Yes. Remember that \( \log 9\), means \( \log_{10} 9\). So, \( \log 9 = \frac{\ln 9}{\ln 10} \).

Example \( \PageIndex{ 14 } \): 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 \(e\).\[ \log_2 10 = \dfrac{\ln 10}{\ln 2} \approx 3.3219 . \nonumber \]

Checkpoint \( \PageIndex{ 14 } \)

Evaluate \( \log_5 (100)\), using the Change-of-Base Formula.