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In this section, you will:
  • 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.
Testing of the pH of hydrochloric acid.
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° 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 a is the concentration of hydrogen ion in the solution

pH = log ( [ H + ] )       = log ( 1 [ H + ] )

The equivalence of log ( [ H + ] ) and log ( 1 [ H + ] ) 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.

log b 1 = 0 log b b = 1

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.

log b ( b x ) = x     b log b x = x , x > 0

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.

log b M = log b N  if and only if   M = N

We can use the one-to-one property to solve the equation log 3 ( 3 x ) = log 3 ( 2 x + 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 :

3 x = 2 x + 5 Set the arguments equal . x = 5 Subtract 2 x .

But what about the equation log 3 ( 3 x ) + log 3 ( 2 x + 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 exponents by adding: 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 1 , we will show

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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