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:
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 + H + is the concentration of hydrogen ion in the solution
pH = − log ( [ H + ] ) = log ( 1 [ H + ] ) pH = − log ( [ H + ] ) = log ( 1 [ H + ] )The equivalence of − log ( [ H + ] ) − log ( [ H + ] ) and log ( 1 [ H + ] ) log ( 1 [ H + ] ) 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. Some important properties of logarithms are given here. First, the following properties are easy to prove.
log b 1 = 0 log b b = 1 log b 1 = 0 log b b = 1For example, log 5 1 = 0 log 5 1 = 0 since 5 0 = 1. 5 0 = 1. And log 5 5 = 1 log 5 5 = 1 since 5 1 = 5. 5 1 = 5.
Next, we have the inverse property.
log b ( b x ) = x b log b x = x , x > 0 log b ( b x ) = x b log b x = x , x > 0For example, to evaluate log ( 100 ) , log ( 100 ) , we can rewrite the logarithm as log 10 ( 10 2 ) , log 10 ( 10 2 ) , and then apply the inverse property log b ( b x ) = x log b ( b x ) = x to get log 10 ( 10 2 ) = 2. log 10 ( 10 2 ) = 2.
To evaluate e ln ( 7 ) , e ln ( 7 ) , we can rewrite the logarithm as e log e 7 , e log e 7 , and then apply the inverse property b log b x = x b log b x = x to get e log e 7 = 7. 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 log b M = log b N if and only if M = NWe can use the one-to-one property to solve the equation log 3 ( 3 x ) = log 3 ( 2 x + 5 ) log 3 ( 3 x ) = log 3 ( 2 x + 5 ) for x . x . Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x : x :
3 x = 2 x + 5 Set the arguments equal . x = 5 Subtract 2 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 ? 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 powers by adding exponents: x a x b = x a + b . 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 x and positive real numbers M , N , M , N , and b , b , where b ≠ 1 , b ≠ 1 , we will show
log b ( M N ) = log b ( M ) + log b ( N ) . log b ( M N ) = log b ( M ) + log b ( N ) .Let m = log b M m = log b M and n = log b N . n = log b N . In exponential form, these equations are b m = M b m = M and b n = N . b n = N . It follows that
log b ( M N ) = log b ( b m b n ) Substitute for M and N . = log b ( b m + n ) Apply the product rule for exponents . = m + n Apply the inverse property of logs . = log b ( M ) + log b ( N ) Substitute for m and n . log b ( M N ) = log b ( b m b n ) Substitute for M and N . = log b ( b m + n ) Apply the product rule for exponents . = m + n Apply the inverse property of logs . = log b ( M ) + log b ( N ) Substitute for m and n .
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 ( w x y z ) . log b ( w x y z ) . Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
log b ( w x y z ) = log b w + log b x + log b y + log b z log b ( w x y z ) = log b w + log b x + log b y + log b z
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.
log b ( M N ) = log b ( M ) + log b ( N ) for b > 0 log b ( M N ) = log b ( M ) + log b ( N ) for b > 0
Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.
Expand log 3 ( 30 x ( 3 x + 4 ) ) . log 3 ( 30 x ( 3 x + 4 ) ) .
We begin by factoring the argument completely, expressing 30 30 as a product of primes.
log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 ⋅ 3 ⋅ 5 ⋅ x ⋅ ( 3 x + 4 ) ) log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 ⋅ 3 ⋅ 5 ⋅ x ⋅ ( 3 x + 4 ) )
Next we write the equivalent equation by summing the logarithms of each factor.
log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 ) + log 3 ( 3 ) + log 3 ( 5 ) + log 3 ( x ) + log 3 ( 3 x + 4 ) log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 ) + log 3 ( 3 ) + log 3 ( 5 ) + log 3 ( x ) + log 3 ( 3 x + 4 )
Expand log b ( 8 k ) . log b ( 8 k ) .
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 a x b = x a − b . 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 x and positive real numbers M , M , N , N , and b , b , where b ≠ 1 , b ≠ 1 , we will show
log b ( M N ) = log b ( M ) − log b ( N ) . log b ( M N ) = log b ( M ) − log b ( N ) .Let m = log b M m = log b M and n = log b N . n = log b N . In exponential form, these equations are b m = M b m = M and b n = N . b n = N . It follows that
log b ( M N ) = log b ( b m b n ) Substitute for M and N . = log b ( b m − n ) Apply the quotient rule for exponents . = m − n Apply the inverse property of logs . = log b ( M ) − log b ( N ) Substitute for m and n . log b ( M N ) = log b ( b m b n ) Substitute for M and N . = log b ( b m − n ) Apply the quotient rule for exponents . = m − n Apply the inverse property of logs . = log b ( M ) − log b ( N ) Substitute for m and n .
For example, to expand log ( 2 x 2 + 6 x 3 x + 9 ) , log ( 2 x 2 + 6 x 3 x + 9 ) , we must first express the quotient in lowest terms. Factoring and canceling we get,
log ( 2 x 2 + 6 x 3 x + 9 ) = log ( 2 x ( x + 3 ) 3 ( x + 3 ) ) Factor the numerator and denominator . = log ( 2 x 3 ) Cancel the common factors . log ( 2 x 2 + 6 x 3 x + 9 ) = log ( 2 x ( x + 3 ) 3 ( x + 3 ) ) Factor the numerator and denominator . = log ( 2 x 3 ) Cancel the common factors .
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.
log ( 2 x 3 ) = log ( 2 x ) − log ( 3 ) = log ( 2 ) + log ( x ) − log ( 3 ) log ( 2 x 3 ) = log ( 2 x ) − log ( 3 ) = log ( 2 ) + log ( x ) − log ( 3 )
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.
log b ( M N ) = log b M − log b N log b ( M N ) = log b M − log b NGiven the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.
Expand log 2 ( 15 x ( x − 1 ) ( 3 x + 4 ) ( 2 − x ) ) . log 2 ( 15 x ( x − 1 ) ( 3 x + 4 ) ( 2 − x ) ) .
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
log 2 ( 15 x ( x − 1 ) ( 3 x + 4 ) ( 2 − x ) ) = log 2 ( 15 x ( x − 1 ) ) − log 2 ( ( 3 x + 4 ) ( 2 − x ) ) log 2 ( 15 x ( x − 1 ) ( 3 x + 4 ) ( 2 − x ) ) = log 2 ( 15 x ( x − 1 ) ) − log 2 ( ( 3 x + 4 ) ( 2 − x ) )
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.
log 2 ( 15 x ( x − 1 ) ) − log 2 ( ( 3 x + 4 ) ( 2 − x ) ) = [ log 2 ( 3 ) + log 2 ( 5 ) + log 2 ( x ) + log 2 ( x − 1 ) ] − [ log 2 ( 3 x + 4 ) + log 2 ( 2 − x ) ] = log 2 ( 3 ) + log 2 ( 5 ) + log 2 ( x ) + log 2 ( x − 1 ) − log 2 ( 3 x + 4 ) − log 2 ( 2 − x ) log 2 ( 15 x ( x − 1 ) ) − log 2 ( ( 3 x + 4 ) ( 2 − x ) ) = [ log 2 ( 3 ) + log 2 ( 5 ) + log 2 ( x ) + log 2 ( x − 1 ) ] − [ log 2 ( 3 x + 4 ) + log 2 ( 2 − x ) ] = log 2 ( 3 ) + log 2 ( 5 ) + log 2 ( x ) + log 2 ( x − 1 ) − log 2 ( 3 x + 4 ) − log 2 ( 2 − x )
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x = − 4 3 x = − 4 3 and x = 2. x = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x > 0 , x > 0 , x > 1 , x > 1 , x > − 4 3 , x > − 4 3 , and x < 2. x < 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.
Expand log 3 ( 7 x 2 + 21 x 7 x ( x − 1 ) ( x − 2 ) ) . log 3 ( 7 x 2 + 21 x 7 x ( x − 1 ) ( x − 2 ) ) .
