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. 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:
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 \([\ce>]\) is the concentration of hydrogen ions in the solution
These two equations for\(pH\) look different but, in fact, are equivalent. This is one of the logarithm properties we will examine in this section.
Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. First, let's recall the properties that we have learned thus far:
The first two properties listed above are called inverse properties.
Recall that we use the product rule of exponents to combine the product of exponents by adding: \(x^ax^b=x^\). 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
Let \(m=<\log>_bM\) and \(n=<\log>_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
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:
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.
We can use this rule as well as others we will discuss shortly, to expand a single logarithm into sums and differences of simpler logarithms or we can use this rule to combine logarithms into a single logarithm. This is a useful things to be able to do as it will allow us a way to solve certain kings of logarithmic equations as well as determine behavior of logarithmic functions. As you will see in calculus, these rules are also very useful for making the process of taking derivatives and integrating much easier. Let's now look at some examples of applying this rule.
Solution
Notice that the argument, \(6a\), is a product. Using the product rule, we can expand to write this as a sum of logarithms, \[\log_5(6a) =\log_5(6)+\log_5(a)\nonumber\]
Remember that is is a common practice to write single term arguments without parentheses. For the last example, it would be acceptable to also write the answer as \(log_56+\log_5a\). However, it would not be correct to write \(\log_5(6a) \text < as >\log_56a\) since the argument is a product. \[\begin\log_56a &=\log_5(6)(a) \qquad \text \\[6pt] &=a\log_56\ \end\]When we have a product of a logarithm with a single term, it is common practice to write the logarithm last.
A common mistake is to not put parentheses around arguments of products. \[\log_b(mn) \neq \log_bmn\nonumber\]
Solution
We begin by factoring the argument completely, expressing \(30\) as a product of primes.
Next we write the equivalent equation by summing the logarithms of each factor.
We discussed earlier that it is acceptable to write a logarithm of a single term argument without parentheses. This is not the case with an argument that is a product or an argument that is a sum or difference.
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: \(x^>=x^\). The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms.
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
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, b, where \(b≠1\), we will show
Let \(m=<\log>_bM\) and \(n=<\log>_bN\). In exponential form, these equations are \(b^m=M\) and \(b^n=N\). It follows that
Solution
Notice that the argument is a quotient. Using the quotient rule are are able to write this is as a difference of logarithms, \[\log_7\left(\dfrac\right)=\log_7(3)-\log_7(x)\nonumber\]
A note about notation here. In the previous example, the argument is a quotient. Notice that the logarithm is aligned with the fraction bar in the argument. Unlike with arguments that are products, sometimes you will see arguments that are quotients written without parentheses, \(\log_7\left(\dfrac\right)=\log_7\dfrac\). Logarithms are functions and functions have inputs. For the statement, \(\log_7\dfrac\), to make sense the argument must be the single term \(\dfrac\).
Solution
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.
Analysis
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for \(x=−\dfrac\) and \(x=2\). Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that \(x>0\), \(x>1\), \(x>−\dfrac\), and \(x
