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Study Guides > Intermediate Algebra

Read: Expand and Condense Logarithms

Learning Objectives

  • Combine product, power and quotient rules to simplify logarithmic expressions
  • Expand logarithmic expressions that have negative or fractional exponents
  • Condense 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:

[latex]\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}[/latex]

We can also 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 (fraction) has a negative power:

[latex]\begin{array}{c}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}[/latex]

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.

traffic-sign-160659-300x265Remember that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Consider the following example:

[latex]\begin{array}{c}\mathrm{log}\left(10+100\right)\overset{?}{=}\end{array}\mathrm{log}\left(10\right)+\mathrm{log}\left(100\right)\\\mathrm{log}\left(110\right)\overset{?}{=}1+2\\2.04\ne3[/latex]

Be careful to only apply the product rule when a logarithm has an argument that is a product or when you have a sum of logarithms.
In our first example we will show that a logarithmic expression can be expanded by combining several of the rules of logarithms.

Example

Rewrite [latex]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)[/latex] as a sum or difference of logs.

Answer:

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

[latex]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)[/latex]

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

[latex]\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex]

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

[latex]\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex]

We can also use the rules for logarithms to simplify the logarithm of a radical expression.

Example

Expand [latex]\mathrm{log}\left(\sqrt{x}\right)[/latex].

Answer: [latex]\begin{array}{c}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}[/latex]

 

Think About it

Can we expand [latex]\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex]? Use the textbox below to develop an argument one way or the other before you look at the solution.[practice-area rows="1"][/practice-area]

Answer: No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Rewrite the expression as an equation and express it as an exponential to give yourself some proof. [latex-display]m=\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex-display] If you rewrite this as an exponential you get: [latex-display]e^m={x}^{2}+{y}^{2}[/latex-display] From here, there's not much more you can do to make this expression more simple.

Let's do one more example with an expression that contains several different mathematical operations.

Example

Expand [latex]{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)[/latex].

Answer:

We can expand by applying the Product and Quotient Rules.

[latex]\begin{array}{c}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient and Product Rules}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}[/latex]

Condense Logarithms

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 [latex]{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)[/latex] as a single logarithm.

Answer:

Using the product and quotient rules

[latex]{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)={\mathrm{log}}_{3}\left(5\cdot 8\right)={\mathrm{log}}_{3}\left(40\right)[/latex]

This reduces our original expression to

[latex]{\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)[/latex]

Then, using the quotient rule

[latex]{\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)={\mathrm{log}}_{3}\left(\frac{40}{2}\right)={\mathrm{log}}_{3}\left(20\right)[/latex]

In our next example, we show how to simplify a more complex logarithm by condensing it.

Example

Condense [latex]{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)[/latex].

Answer:

We apply the power rule first:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex]

Next we apply the product rule to the sum:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex]

Finally, we apply the quotient rule to the difference:

[latex]{\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}[/latex]

 Summary

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.

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  • Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..