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# Expanding and Condensing Logarithms

### Learning Outcomes

• Expand a logarithm using a combination of logarithm rules.
• Condense a logarithmic expression into one logarithm.

## Expanding Logarithms

Taken together, the product rule, quotient rule, and power rule are often called "properties of logs." Sometimes we apply more than one rule in order to expand an expression. For example:

$\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}$

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:

$\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}$

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 and then just 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: Using a combination of the rules for logarithms to expand a logarithm

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

Answer: First, because we have a quotient of two expressions, we can use the quotient rule: [latex-display]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)[/latex-display] Then seeing the product in the first term, we use the product rule: [latex-display]\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-display] Finally, we use the power rule on the first term: [latex-display]\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-display]

### Tip for success

When expanding and condensing logarithms, keep in mind that there are often more than one or two good ways to reach a good conclusion. The rules for manipulating exponents and logarithms can be combined creatively. You should try a few different ideas for using the rules on complicated expressions to get practice for finding the most efficient path to take in different situations.

### Try It

Expand $\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right)$.

[ohm_question]35034[/ohm_question]
In the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.

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

Expand $\mathrm{log}\left(\sqrt{x}\right)$.

Answer: [latex-display]\begin{array}{l}\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-display]

### Try It

Expand $\mathrm{ln}\left(\sqrt[3]{{x}^{2}}\right)$.

Answer: $\frac{2}{3}\mathrm{ln}x$

[ohm_question]129752[/ohm_question]

### Q & A

Can we expand $\mathrm{ln}\left({x}^{2}+{y}^{2}\right)$? No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Now we will provide some examples that will require careful attention.

### tip for success

When working with more complicated examples, write your work down step by step to avoid making incorrect assumptions. It's okay to try different rules as you practice creativity as long as you use each rule correctly.

### Example: Expanding Complex Logarithmic Expressions

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

Answer: We can expand by applying the product and quotient rules. [latex-display]\begin{array}{lllllll}{\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 product and quotient rule}.\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-display]

### Try It

Expand $\mathrm{ln}\left(\frac{\sqrt{\left(x - 1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right)$.

Answer: $\frac{1}{2}\mathrm{ln}\left(x - 1\right)+\mathrm{ln}\left(2x+1\right)-\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x - 3\right)$

[ohm_question]129764[/ohm_question]

## Condensing 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.

### 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. From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.

### Example: Using the Power Rule in Reverse

Use the power rule for logs to rewrite $4\mathrm{ln}\left(x\right)$ as a single logarithm with a leading coefficient of 1.

Answer: 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 $4\mathrm{ln}\left(x\right)$, we identify the factor, 4, as the exponent and the argument, x, as the base and rewrite the product as a logarithm of a power: $4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right)$.

### Try It

Use the power rule for logs to rewrite $2{\mathrm{log}}_{3}4$ as a single logarithm with a leading coefficient of 1.

Answer: ${\mathrm{log}}_{3}16$

In our next few examples we will use a combination of logarithm rules to condense logarithms.

### Example: Using the Product and Quotient Rules to Combine Logarithms

Write ${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)$ as a single logarithm.

Answer: From left to right, since we have the addition of two logs, we first use the product rule: [latex-display]{\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-display] This simplifies our original expression to: [latex-display]{\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)[/latex-display] Using the quotient rule: [latex-display]{\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-display]

### Try It

Condense $\mathrm{log}3-\mathrm{log}4+\mathrm{log}5-\mathrm{log}6$.

Answer: $\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right)$; can also be written $\mathrm{log}\left(\frac{5}{8}\right)$ by simplifying the fraction to lowest terms.

[ohm_question]129766[/ohm_question]

### Example: Condensing Complex Logarithmic Expressions

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

Answer: We apply the power rule first: [latex-display]{\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-display] From left to right, since we have the addition of two logs, we apply the product rule to the sum: [latex-display]{\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-display] Finally we apply the quotient rule to the difference: [latex-display]{\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-display]

### Example: Rewriting as a Single Logarithm

Rewrite $2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)$ as a single logarithm.

Answer: We apply the power rule first: [latex-display]2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)[/latex-display] From left to right, since we have the difference of two logs, we apply the quotient rule to the difference: [latex-display]\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)[/latex-display] Finally we apply the product rule to the sum: [latex-display]\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{{x}^{2}}{{\left(3x+5\right)}^{{x}^{-1}}}}{{\left(x+5\right)}^{4}}\right)[/latex-display]

### Try It

Rewrite $\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x - 1\right)+3\mathrm{log}\left(x - 1\right)$ as a single logarithm.

Answer: $\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)$

Condense $4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right)$.

Answer: $\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}}$; this answer could also be written as $\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}$.

[ohm_question]129768[/ohm_question]

## Applications of Properties of Logarithms

In chemistry, pH is a measure of how acidic or basic a liquid is. It is essentially a measure of the concentration of hydrogen ions in a solution. The scale for measuring pH is standardized across the world, the scientific community having agreed upon its values and methods for acquiring them. Measurements of pH can help scientists, farmers, doctors, and engineers solve problems and identify sources of problems.

pH is defined as the decimal logarithm of the reciprocal of the hydrogen ion activity, $a_{H}+$, in a solution. $\text{pH} =-\log _{10}(a_{{\text{H}}^{+}})=\log _{10}\left({\frac {1}{a_{{\text{H}}^{+}}}}\right)$

For example, a solution with a hydrogen ion activity of $2.5×{10}^{-6}$ (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of $\log_{10}\left(\frac{1}{2.5×{10}^{-6}}\right)=5.6$

In the next examples, we will solve some problems involving pH.

### Example: Applying Properties of Logs

Recall that, in chemistry, $\text{pH}=-\mathrm{log}\left[{H}^{+}\right]$. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Answer: Suppose C is the original concentration of hydrogen ions and P is the original pH of the liquid. Then $\text{P}=-\mathrm{log}\left(C\right)$. If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is $\text{pH}=-\mathrm{log}\left(2C\right)$ Using the product rule of logs [latex-display]\text{pH}=-\mathrm{log}\left(2C\right)=-\left(\mathrm{log}\left(2\right)+\mathrm{log}\left(C\right)\right)=-\mathrm{log}\left(2\right)-\mathrm{log}\left(C\right)[/latex-display] Since $P=-\mathrm{log}\left(C\right)$, the new pH is [latex-display]\text{pH}=P-\mathrm{log}\left(2\right)\approx P - 0.301[/latex-display] When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

### Try It

How does the pH change when the concentration of positive hydrogen ions is decreased by half?