Logarithmic Properties

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 basic. 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 basic, 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° C)
• Human blood: 7.35
• Fresh coconut: 7.8
• Sodium hydroxide (lye): 14

To determine whether a solution is acidic or basic, 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.

The equivalence of $-\log\left([H^+]\right)$ and $\log\left(\tfrac{1}{[H^+]}\right)$ is one of the logarithm properties we will examine in this section.

Using the Product Rule for Logarithms

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. First, the following properties are easy to prove.

For example, $\log_5 1=0$ since $5^0=1$. And $\log_5 5=1$ since $5^1=5$.

Next, we have the inverse property.

For example, to evaluate $\log(100)$, we can rewrite the logarithm as $\log_{10}(10^2)$, and then apply the inverse property $\log_b(b^x)=x$ 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 property $b^{\log_b x}=x$ to get $e^{\log_e 7}=7$.

Finally, we have the one-to-one property.

We can use the one-to-one property 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 property by setting the arguments equal and solving for $x$:

But what about the equation $\log_3(3x)+\log_3(2x+5)=2$? The one-to-one property 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 rule of exponents to combine the product of powers by adding exponents: $x^a x^b=x^{a+b}$. 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 $b\ne1$, we will show

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

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:

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

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.

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: $\tfrac{x^a}{x^b}=x^{a-b}$. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of 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 $b$, where $b\ne1$, we will show

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

For example, to expand $\log\left(\tfrac{2x^2+6x}{3x+9}\right)$, we must first express the quotient in lowest terms. Factoring and canceling we get,

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

Example. Expand $\log_2\left(\tfrac{15x(x-1)}{(3x+4)(2-x)}\right)$.

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 $x=-\tfrac{4}{3}$ and $x=2$. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that $x>0$, $x>1$, $x>-\tfrac{4}{3}$, and $x<2$. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

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:

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 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,

Example. 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.

Example. Expand $\log_3(25)$ using the power rule 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.

Example. Rewrite $4\ln(x)$ using the power rule 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: $4\ln(x)=\ln(x^4)$.

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:

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:

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 products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.

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

Solution. First, because we have a quotient of two expressions, we can use the quotient rule:

Then seeing the product in the first term, we use the product rule:

Finally, we use the power rule on the first term:

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

Solution.

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

Solution. We can expand by applying the Product and Quotient Rules.

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.

Example. Write $\log_3(5)+\log_3(8)-\log_3(2)$ as a single logarithm.

Solution. Using the product and quotient rules

This reduces our original expression to

Then, using the quotient rule

Example. Condense $\log_2(x^2)+\tfrac12\log_2(x-1)-3\log_2((x+3)^2)$.

Solution. We apply the power rule first:

Next we apply the product rule to the sum:

Finally, we apply the quotient rule to the difference:

Example. Rewrite $2\log x-4\log(x+5)+\tfrac1x\log(3x+5)$ as a single logarithm.

Solution. We apply the power rule first:

Next we rearrange and apply the product rule to the sum:

Finally, we apply the quotient rule to the difference:

Example. 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

Using the product rule of logs

Since $P=-\log(C)$, the new pH is

When the concentration of hydrogen ions is doubled, the pH decreases by about $0.301$.

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.

To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms.

Given any positive real numbers $M,b,$ and $n$, where $n\ne1$ and $b\ne1$, we show

Let $y=\log_b M$. By exponentiating both sides with base $b$, we arrive at an exponential form, namely $b^y=M$. It follows that

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.

Example. 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.

Example. 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 $e$.

Key equations

Key concepts

• We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.
• We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.
• 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.
• We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.
• The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.
• We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.
• 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.

Practice

Use the product rule for logarithms

Use the quotient rule for logarithms

Use the power rule for logarithms

Expand logarithmic expressions

Condense logarithmic expressions

Use the change-of-base formula for logarithms

This section is adapted from Precalculus 2e, Section 4.5: Logarithmic Properties by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the pinned CNXML’s “coreq-skills” block, a prerequisite/corequisite review section carrying its own “Learning Objectives” (exponent-property and logarithm-property review keyed to Intermediate Algebra sections 5.2.1 and 10.4.1) that precedes the real section content in the source file but is not part of the printed textbook — the rendered PDF (pages 470–471) runs directly from the end of Section 4.4 to this section’s own six-item Learning Objectives list and its pH introduction, with no exponent-review material printed anywhere on those pages; omitted the decorative photograph of litmus paper testing hydrochloric acid, which carries no mathematics; converted every source Try It into a real interactive component; five Try Its (the quotient-rule expansion of $\log_3\left(\tfrac{7x^2+21x}{7x(x-1)(x-2)}\right)$; the power-rule expansions of $\ln x^2$ and $\ln\left(\tfrac{1}{x^2}\right)$; the root expansion $\ln\left(\sqrt[3]{x^2}\right)$; and the complex expansion $\ln\left(\tfrac{\sqrt{(x-1)(2x+1)^2}}{x^2-9}\right)$) and one end-of-section exercise (expanding $\ln\left(\tfrac{1}{4^k}\right)$) became multiple-choice questions rather than fill-ins, because their fully-expanded answers either keep an irreducible binomial inside a logarithm’s argument (the expanded-logarithms answer form accepts only a single number or variable there) or are written with $\ln$, which the pinned answer-form checker’s expanded-logarithms predicate does not scan for at all, so a fill-in declaring that form would silently accept the unexpanded printed expression retyped back; verified independently that this is a grader limitation, not a content issue, before adapting each one; the pH-decrease-by-half Try It and the evaluate-$\log_5(100)$ Try It ask for a rounded decimal rather than an exact log expression, matching the pinned solution’s own three-decimal rounding; and adapted fourteen selected end-of-section exercises — two product-rule, two quotient-rule, two power-rule, three general expansions, two general condensations, and three change-of-base evaluations — into interactive components in a closing Practice block, one group per objective. The Practice item rewriting $-\log_b\left(\tfrac17\right)$ is worded “use the Power Rule to rewrite… as a single logarithm” rather than the source’s “condense to a single logarithm,” and declares expanded-logarithms rather than single-logarithm: the printed prompt is itself already a single logarithm (negated), so single-logarithm — which accepts a leading unary minus as part of “the term IS the logarithm” — would accept it retyped verbatim; expanded-logarithms’s single-atomic-argument requirement correctly rejects the un-simplified $\tfrac17$ argument while accepting the simplified $\log_b(7)$. The two “quotient of logs” change-of-base items ($\log_{0.5}(8)$ and $\log_7(15)$) declare single-fraction rather than the seemingly natural exact-log: exact-log’s own grammar admits a bare, unconverted logarithm as a zero-integer-coefficient case, so it would accept the printed logarithm retyped back without the required change-of-base conversion; single-fraction correctly requires the one-quotient shape the conversion produces.