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Guias de estudo > College Algebra

Logarithm Rules

Learning Objectives

  • Rewrite a logarithmic expression using the power rule, product rule, or quotient rule
  • Expand logarithmic expressions using a combination of logarithm rules
  • Condense logarithmic expressions using logarithm rules
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=1\begin{array}{l}{\mathrm{log}}_{b}1=0\\{\mathrm{log}}_{b}b=1\end{array}

For example, log51=0{\mathrm{log}}_{5}1=0 since 50=1{5}^{0}=1. And log55=1{\mathrm{log}}_{5}5=1 since 51=5{5}^{1}=5. Next, we have the inverse property.

logb(bx)=x blogbx=x,x>0\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}

For example, to evaluate log(100)\mathrm{log}\left(100\right), we can rewrite the logarithm as log10(102){\mathrm{log}}_{10}\left({10}^{2}\right), and then apply the inverse property logb(bx)=x{\mathrm{log}}_{b}\left({b}^{x}\right)=x to get log10(102)=2{\mathrm{log}}_{10}\left({10}^{2}\right)=2. To evaluate eln(7){e}^{\mathrm{ln}\left(7\right)}, we can rewrite the logarithm as eloge7{e}^{{\mathrm{log}}_{e}7}, and then apply the inverse property blogbx=x{b}^{{\mathrm{log}}_{b}x}=x to get eloge7=7{e}^{{\mathrm{log}}_{e}7}=7. Finally, we have the one-to-one property.

logbM=logbN if and only if M=N{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N

We can use the one-to-one property to solve the equation log3(3x)=log3(2x+5){\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right) 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:

3x=2x+5Set the arguments equal.x=5Subtract 2x.\begin{array}{l}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}

But what about the equation log3(3x)+log3(2x+5)=2{\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=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: xaxb=xa+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 and positive real numbers M, N, and b, where b≠1b\ne 1, we will show

logb(MN)=logb(M)+logb(N){\mathrm{log}}_{b}\left(MN\right)\text{=}{\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right).

Let m=logbMm={\mathrm{log}}_{b}M and n=logbNn={\mathrm{log}}_{b}N. In exponential form, these equations are bm=M{b}^{m}=M and bn=N{b}^{n}=N. It follows that

logb(MN)=logb(bmbn)Substitute for M and N.=logb(bm+n)Apply the product rule for exponents.=m+nApply the inverse property of logs.=logb(M)+logb(N)Substitute for m and n.\begin{array}{l}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \\ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}

 

A General Note: 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>0{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0

Example: Using the Product Rule for Logarithms

Expand log3(30x(3x+4)){\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right).

Answer: We begin by writing the equivalent equation by summing the logarithms of each factor.

log3(30x(3x+4))=log3(30x)+log3(3x+4)=log3(30)+log3(x)+log3(3x+4){\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(30x\right)+{\mathrm{log}}_{3}\left(3x+4\right)={\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)

The final expansion looks like this, note how the factor 30x30x can be expanded into the sum of two logarithms:

log3(30)+log3(x)+log3(3x+4){\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)

Try It

Expand logb(8k){\mathrm{log}}_{b}\left(8k\right).

Answer: logb8+logbk{\mathrm{log}}_{b}8+{\mathrm{log}}_{b}k

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{x}^{2}? One method is as follows:

logb(x2)=logb(x⋅x)=logbx+logbx=2logbx\begin{array}{l}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}

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−1\begin{array}{l}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}

A General Note: 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)=nlogbM{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M

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: Expanding a Logarithm with Powers

Expand log2x5{\mathrm{log}}_{2}{x}^{5}.

Answer: 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.

log2(x5)=5log2x{\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x

Try It

Expand lnx2\mathrm{ln}{x}^{2}.

Answer: 2lnx2\mathrm{ln}x

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