Logarithmic Properties

Learning Objectives

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.

Figure 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 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+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.

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.

logb1=0logbb=1log𝑏1=0log𝑏𝑏=1

For example, log51=0log51=0 since 50=1.50=1. And log55=1log55=1 since 51=5.51=5.

Next, we have the inverse property.

logb(bx)=x  blogbx=x,x>0log𝑏(𝑏𝑥)=𝑥  𝑏log𝑏𝑥=𝑥,𝑥>0

For example, to evaluate log(100),log(100), we can rewrite the logarithm as log10(102),log10(102), and then apply the inverse property logb(bx)=xlog𝑏(𝑏𝑥)=𝑥 to get log10(102)=2.log10(102)=2.

To evaluate eln(7),𝑒ln(7), we can rewrite the logarithm as eloge7,𝑒log𝑒7, and then apply the inverse property blogbx=x𝑏log𝑏𝑥=𝑥 to get eloge7=7.𝑒log𝑒7=7.

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

logbM=logbNif and only ifM=Nlog𝑏𝑀=log𝑏𝑁if and only if𝑀=𝑁

We can use the one-to-one property to solve the equation log3(3x)=log3(2x+5)log3(3𝑥)=log3(2𝑥+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:

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

But what about the equation log3(3x)+log3(2x+5)=2?log3(3𝑥)+log3(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: xaxb=xa+b.𝑥𝑎𝑥𝑏=𝑥𝑎+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 x𝑥 and positive real numbers M,N,𝑀,𝑁, and b,𝑏, where b≠1,𝑏≠1, we will show

logb(MN)=logb(M)+logb(N).log𝑏(𝑀𝑁)=log𝑏(𝑀)+log𝑏(𝑁).

Let m=logbM𝑚=log𝑏𝑀 and n=logbN.𝑛=log𝑏𝑁. In exponential form, these equations are bm=M𝑏𝑚=𝑀 and bn=N.𝑏𝑛=𝑁. It follows that

logb(MN)=logb(bmbn)=logb(bm+n)=m+n=logb(M)+logb(N)Substitute for Mand N.Apply the product rule for exponents.Apply the inverse property of logs.Substitute for mand n.log𝑏(𝑀𝑁)=log𝑏(𝑏𝑚𝑏𝑛)Substitute for 𝑀and 𝑁.=log𝑏(𝑏𝑚+𝑛)Apply the product rule for exponents.=𝑚+𝑛Apply the inverse property of logs.=log𝑏(𝑀)+log𝑏(𝑁)Substitute for 𝑚and 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 logb(wxyz).log𝑏(𝑤𝑥𝑦𝑧). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

logb(wxyz)=logbw+logbx+logby+logbzlog𝑏(𝑤𝑥𝑦𝑧)=log𝑏𝑤+log𝑏𝑥+log𝑏𝑦+log𝑏𝑧

THE PRODUCT RULE FOR LOGARITHMS

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

logb(MN)=logb(M)+logb(N)for b>0log𝑏(𝑀𝑁)=log𝑏(𝑀)+log𝑏(𝑁)for 𝑏>0

HOW TO

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

1. Factor the argument completely, expressing each whole number factor as a product of primes.
2. Write the equivalent expression by summing the logarithms of each factor.

EXAMPLE 1

Using the Product Rule for Logarithms

Expand log3(30x(3x+4)).log3(30𝑥(3𝑥+4)).

Solution

We begin by factoring the argument completely, expressing 3030 as a product of primes.

log3(30x(3x+4))=log3(2⋅3⋅5⋅x⋅(3x+4))log3(30𝑥(3𝑥+4))=log3(2⋅3⋅5⋅𝑥⋅(3𝑥+4))

Next we write the equivalent equation by summing the logarithms of each factor.

log3(30x(3x+4))=log3(2)+log3(3)+log3(5)+log3(x)+log3(3x+4)log3(30𝑥(3𝑥+4))=log3(2)+log3(3)+log3(5)+log3(𝑥)+log3(3𝑥+4)

TRY IT #1

Expand logb(8k).log𝑏(8𝑘).

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: xaxb=xa−b.𝑥𝑎𝑥𝑏=𝑥𝑎−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 x𝑥 and positive real numbers M,𝑀, N,𝑁, and b,𝑏, where b≠1,𝑏≠1, we will show

logb(MN)=logb(M)−logb(N).log𝑏(𝑀𝑁)=log𝑏(𝑀)−log𝑏(𝑁).

Let m=logbM𝑚=log𝑏𝑀 and n=logbN.𝑛=log𝑏𝑁. In exponential form, these equations are bm=M𝑏𝑚=𝑀 and bn=N.𝑏𝑛=𝑁. It follows that

logb(MN)=logb(bmbn)=logb(bm−n)=m−n=logb(M)−logb(N)Substitute for Mand N.Apply the quotient rule for exponents.Apply the inverse property of logs.Substitute for mand n.log𝑏(𝑀𝑁)=log𝑏(𝑏𝑚𝑏𝑛)Substitute for 𝑀and 𝑁.=log𝑏(𝑏𝑚−𝑛)Apply the quotient rule for exponents.=𝑚−𝑛Apply the inverse property of logs.=log𝑏(𝑀)−log𝑏(𝑁)Substitute for 𝑚and 𝑛.

For example, to expand log(2×2+6x3x+9),log(2𝑥2+6𝑥3𝑥+9), we must first express the quotient in lowest terms. Factoring and canceling we get,

log(2×2+6x3x+9)=log(2x(x+3)3(x+3))                      =log(2×3)Factor the numerator and denominator.Cancel the common factors.log(2𝑥2+6𝑥3𝑥+9)=log(2𝑥(𝑥+3)3(𝑥+3))Factor the numerator and denominator.                      =log(2𝑥3)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(2×3)=log(2x)−log(3)            =log(2)+log(x)−log(3)log(2𝑥3)=log(2𝑥)−log(3)            =log(2)+log(𝑥)−log(3)

THE QUOTIENT RULE FOR 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.

logb(MN)=logbM−logbNlog𝑏(𝑀𝑁)=log𝑏𝑀−log𝑏𝑁

HOW TO

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

EXAMPLE 2

Using the Quotient Rule for Logarithms

Expand log2(15x(x−1)(3x+4)(2−x)).log2(15𝑥(𝑥−1)(3𝑥+4)(2−𝑥)).

Solution

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

log2(15x(x−1)(3x+4)(2−x))=log2(15x(x−1))−log2((3x+4)(2−x))log2(15𝑥(𝑥−1)(3𝑥+4)(2−𝑥))=log2(15𝑥(𝑥−1))−log2((3𝑥+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.

log2(15x(x−1))−log2((3x+4)(2−x))=[log2(3)+log2(5)+log2(x)+log2(x−1)]−[log2(3x+4)+log2(2−x)]                                                                =log2(3)+log2(5)+log2(x)+log2(x−1)−log2(3x+4)−log2(2−x)log2(15𝑥(𝑥−1))−log2((3𝑥+4)(2−𝑥))=[log2(3)+log2(5)+log2(𝑥)+log2(𝑥−1)]−[log2(3𝑥+4)+log2(2−𝑥)]                                                                =log2(3)+log2(5)+log2(𝑥)+log2(𝑥−1)−log2(3𝑥+4)−log2(2−𝑥)

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=−43𝑥=−43 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>−43,𝑥>−43, 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.

TRY IT #2

Expand log3(7×2+21x7x(x−1)(x−2)).log3(7𝑥2+21𝑥7𝑥(𝑥−1)(𝑥−2)).

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 x2?𝑥2? One method is as follows:

logb(x2)=logb(x⋅x)=logbx+logbx=2logbxlog𝑏(𝑥2)=log𝑏(𝑥⋅𝑥)=log𝑏𝑥+log𝑏𝑥=2log𝑏𝑥

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,

100=1023–√=3121e=e−1100=1023=3121𝑒=𝑒−1

THE POWER RULE FOR LOGARITHMS

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

logb(Mn)=nlogbMlog𝑏(𝑀𝑛)=𝑛log𝑏𝑀

HOW TO

Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.

1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.

EXAMPLE 3

Expanding a Logarithm with Powers

Expand log2x5.log2𝑥5.

Solution

The argument is already written as a power, so we identify the exponent, 5, and the base, x,𝑥, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

log2(x5)=5log2xlog2(𝑥5)=5log2𝑥

TRY IT #3

Expand lnx2.ln𝑥2.

EXAMPLE 4

Rewriting an Expression as a Power before Using the Power Rule

Expand log3(25)log3(25) using the power rule for logs.

Solution

Expressing the argument as a power, we get log3(25)=log3(52).log3(25)=log3(52).

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.

log3(52)=2log3(5)log3(52)=2log3(5)

TRY IT #4

Expand ln(1×2).ln(1𝑥2).

EXAMPLE 5

Using the Power Rule in Reverse

Rewrite 4ln(x)4ln(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 4ln(x),4ln(𝑥), we identify the factor, 4, as the exponent and the argument, x,𝑥, as the base, and rewrite the product as a logarithm of a power: 4ln(x)=ln(x4).4ln(𝑥)=ln(𝑥4).

TRY IT #5

Rewrite 2log342log34 using the power rule for logs to a single logarithm with a leading coefficient of 1.

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:

logb(6xy)=logb(6x)−logby=logb6+logbx−logbylog𝑏(6𝑥𝑦)=log𝑏(6𝑥)−log𝑏𝑦=log𝑏6+log𝑏𝑥−log𝑏𝑦

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:

logb(AC)=logb(AC−1)=logb(A)+logb(C−1)=logbA+(−1)logbC=logbA−logbClog𝑏(𝐴𝐶)=log𝑏(𝐴𝐶−1)=log𝑏(𝐴)+log𝑏(𝐶−1)=log𝑏𝐴+(−1)log𝑏𝐶=log𝑏𝐴−log𝑏𝐶

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 6

Expanding Logarithms Using Product, Quotient, and Power Rules

Rewrite ln(x4y7)ln(𝑥4𝑦7) as a sum or difference of logs.

Solution

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

ln(x4y7)=ln(x4y)−ln(7)ln(𝑥4𝑦7)=ln(𝑥4𝑦)−ln(7)

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

ln(x4y)−ln(7)=ln(x4)+ln(y)−ln(7)ln(𝑥4𝑦)−ln(7)=ln(𝑥4)+ln(𝑦)−ln(7)

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

ln(x4)+ln(y)−ln(7)=4ln(x)+ln(y)−ln(7)ln(𝑥4)+ln(𝑦)−ln(7)=4ln(𝑥)+ln(𝑦)−ln(7)

TRY IT #6

Expand log(x2y3z4).log(𝑥2𝑦3𝑧4).

EXAMPLE 7

Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression

Expand log(x−−√).log(𝑥).

Solution

log(x−−√)=logx(12)=12logxlog(𝑥)=log𝑥(12)=12log𝑥

TRY IT #7

Expand ln(x2−−√3).ln(𝑥23).

Q&A

Can we expand ln(x2+y2)?ln(𝑥2+𝑦2)?

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.

EXAMPLE 8

Expanding Complex Logarithmic Expressions

Expand log6(64×3(4x+1)(2x−1)).log6(64𝑥3(4𝑥+1)(2𝑥−1)).

Solution

We can expand by applying the Product and Quotient Rules.

log6(64×3(4x+1)(2x−1))=log664+log6x3+log6(4x+1)−log6(2x−1)=log626+log6x3+log6(4x+1)−log6(2x−1)=6log62+3log6x+log6(4x+1)−log6(2x−1)Apply the Quotient Rule.Simplify by writing  64 as 26.Apply the Power Rule.log6(64𝑥3(4𝑥+1)(2𝑥−1))=log664+log6𝑥3+log6(4𝑥+1)−log6(2𝑥−1)Apply the Quotient Rule.=log626+log6𝑥3+log6(4𝑥+1)−log6(2𝑥−1)Simplify by writing  64 as 26.=6log62+3log6𝑥+log6(4𝑥+1)−log6(2𝑥−1)Apply the Power Rule.

TRY IT #8

Expand ln((x−1)(2x+1)2√(x2−9)).ln((𝑥−1)(2𝑥+1)2(𝑥2−9)).

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.

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 property 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 property. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.

EXAMPLE 9

Using the Product and Quotient Rules to Combine Logarithms

Write log3(5)+log3(8)−log3(2)log3(5)+log3(8)−log3(2) as a single logarithm.

Solution

Using the product and quotient rules

log3(5)+log3(8)=log3(5⋅8)=log3(40)log3(5)+log3(8)=log3(5⋅8)=log3(40)

This reduces our original expression to

log3(40)−log3(2)log3(40)−log3(2)

Then, using the quotient rule

log3(40)−log3(2)=log3(402)=log3(20)log3(40)−log3(2)=log3(402)=log3(20)

TRY IT #9

Condense log3−log4+log5−log6.log3−log4+log5−log6.

EXAMPLE 10

Condensing Complex Logarithmic Expressions

Condense log2(x2)+12log2(x−1)−3log2((x+3)2).log2(𝑥2)+12log2(𝑥−1)−3log2((𝑥+3)2).

Solution

We apply the power rule first:

log2(x2)+12log2(x−1)−3log2((x+3)2)=log2(x2)+log2(x−1−−−−√)−log2((x+3)6)log2(𝑥2)+12log2(𝑥−1)−3log2((𝑥+3)2)=log2(𝑥2)+log2(𝑥−1)−log2((𝑥+3)6)

Next we apply the product rule to the sum:

log2(x2)+log2(x−1−−−−√)−log2((x+3)6)=log2(x2x−1−−−−√)−log2((x+3)6)log2(𝑥2)+log2(𝑥−1)−log2((𝑥+3)6)=log2(𝑥2𝑥−1)−log2((𝑥+3)6)

Finally, we apply the quotient rule to the difference:

log2(x2x−1−−−−√)−log2((x+3)6)=log2x2x−1−−−−√(x+3)6log2(𝑥2𝑥−1)−log2((𝑥+3)6)=log2𝑥2𝑥−1(𝑥+3)6

TRY IT #10

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

EXAMPLE 11

Rewriting as a Single Logarithm

Rewrite 2logx−4log(x+5)+1xlog(3x+5)2log𝑥−4log(𝑥+5)+1𝑥log(3𝑥+5) as a single logarithm.

Solution

We apply the power rule first:

2logx–4log(x+5)+1xlog(3x+5)=log(x2)−log(x+5)4+log((3x+5)x−1)2log𝑥–4log(𝑥+5)+1𝑥log(3𝑥+5)=log(𝑥2)−log(𝑥+5)4+log((3𝑥+5)𝑥−1)

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

log(x2)−log(x+5)4+log((3x+5)x−1)log(𝑥2)−log(𝑥+5)4+log((3𝑥+5)𝑥−1)

=log(x2)+log((3x+5)x−1)−log(x+5)4=log(𝑥2)+log((3𝑥+5)𝑥−1)−log(𝑥+5)4

=log(x2(3x+5)x−1)−log(x+5)4=log(𝑥2(3𝑥+5)𝑥−1)−log(𝑥+5)4

Finally, we apply the quotient rule to the difference:

=log(x2(3x+5)x−1)−log(x+5)4=logx2(3x+5)x−1(x+5)4=log(𝑥2(3𝑥+5)𝑥−1)−log(𝑥+5)4=log𝑥2(3𝑥+5)𝑥−1(𝑥+5)4

TRY IT #11

Condense 4(3log(x)+log(x+5)−log(2x+3)).4(3log(𝑥)+log(𝑥+5)−log(2𝑥+3)).

EXAMPLE 12

Applying of the Laws of Logs

Recall that, in chemistry, pH=−log[H+].pH=−log[𝐻+]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Solution

Suppose C𝐶 is the original concentration of hydrogen ions, and P𝑃 is the original pH of the liquid. Then P=–log(C).𝑃=–log(𝐶). If the concentration is doubled, the new concentration is 2C.2𝐶. Then the pH of the new liquid is

pH=−log(2C)pH=−log(2𝐶)

Using the product rule of logs

pH=−log(2C)=−(log(2)+log(C))=−log(2)−log(C)pH=−log(2𝐶)=−(log(2)+log(𝐶))=−log(2)−log(𝐶)

Since P=–log(C),𝑃=–log(𝐶), the new pH is

pH=P−log(2)≈P−0.301pH=𝑃−log(2)≈𝑃−0.301

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

TRY IT #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,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,𝑀,𝑏, and n,𝑛, where n≠1 𝑛≠1  and b≠1,𝑏≠1, we show

logbM=lognMlognblog𝑏𝑀=log𝑛𝑀log𝑛𝑏

Let y=logbM.𝑦=log𝑏𝑀.By exponentiating both sides with baseb𝑏, we arrive at an exponential form, namely by=M.𝑏𝑦=𝑀. It follows that

logn(by)ylognbylogbM=lognM=lognM =lognMlognb=lognMlognbApply the one-to-one property.Apply the power rule for logarithms.Isolate y.Substitute for y.log𝑛(𝑏𝑦)=log𝑛𝑀Apply the one-to-one property.𝑦log𝑛𝑏=log𝑛𝑀 Apply the power rule for logarithms.𝑦=log𝑛𝑀log𝑛𝑏Isolate 𝑦.log𝑏𝑀=log𝑛𝑀log𝑛𝑏Substitute for 𝑦.

For example, to evaluate log536log536 using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

log536=log(36)log(5)≈2.2266Apply the change of base formula using base 10.Use a calculator to evaluate to 4 decimal places.log536=log(36)log(5)Apply the change of base formula using base 10.≈2.2266Use a calculator to evaluate to 4 decimal places.

THE CHANGE-OF-BASE FORMULA

The change-of-base formula can be used to evaluate a logarithm with any base.

For any positive real numbers M,b,𝑀,𝑏, and n,𝑛, where n≠1 𝑛≠1  and b≠1,𝑏≠1,

logbM=lognMlognb.log𝑏𝑀=log𝑛𝑀log𝑛𝑏.

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

logbM=lnMlnblog𝑏𝑀=ln𝑀ln𝑏

and

logbM=logMlogblog𝑏𝑀=log𝑀log𝑏

HOW TO

Given a logarithm with the form logbM,log𝑏𝑀, use the change-of-base formula to rewrite it as a quotient of logs with any positive base n,𝑛, where n≠1.𝑛≠1.

1. Determine the new base n,𝑛, remembering that the common log, log(x),log(𝑥), has base 10, and the natural log, ln(x),ln(𝑥), has base e.𝑒.
2. Rewrite the log as a quotient using the change-of-base formula
   1. The numerator of the quotient will be a logarithm with base n𝑛 and argument M.𝑀.
   2. The denominator of the quotient will be a logarithm with base n𝑛 and argument b.𝑏.

EXAMPLE 13

Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Change log53log53 to a quotient of natural logarithms.

Solution

Because we will be expressing log53log53 as a quotient of natural logarithms, the new base, n=e.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.

logbMlog53=lnMlnb=ln3ln5log𝑏𝑀=ln𝑀ln𝑏log53=ln3ln5

TRY IT #13

Change log0.58log0.58 to a quotient of natural logarithms.

Q&A

Can we change common logarithms to natural logarithms?

Yes. Remember that log9log9 means log109.log109. So, log9=ln9ln10.log9=ln9ln10.

EXAMPLE 14

Using the Change-of-Base Formula with a Calculator

Evaluate log2(10)log2(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.𝑒.

log210=ln10ln2≈3.3219Apply the change of base formula using base e.Use a calculator to evaluate to 4 decimal places.log210=ln10ln2Apply the change of base formula using base 𝑒.≈3.3219Use a calculator to evaluate to 4 decimal places.

TRY IT #14

Evaluate log5(100)log5(100) using the change-of-base formula.