Expand and Condense Logarithms

Learning Objectives

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

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:

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

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, 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.

Solution

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

$\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)$

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

$\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$

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

$\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)$

Try It

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

Solution

$2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z$

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

Solution

$\begin{array}{l}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}$

Try It

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

Solution

\frac{2}{3} l n x

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

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.

Example: Expanding Complex Logarithmic Expressions

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

Solution

We can expand by applying the Product and Quotient Rules.

$\begin{array}{l}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient Rule}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}$

Try It 8

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

Solution

\frac{1}{2} l n (x - 1) + l n (2 x + 1) - l n (x + 3) - l n (x - 3)

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

Condense 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.

Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.

Example: Using the Power Rule in Reverse

Rewrite $4\mathrm{ln}\left(x\right)$ 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 $4\mathrm{ln}\left(x\right)$ , 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

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

Solution

{l o g}_{3} 16

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

In our next 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.

Solution

Using the product and quotient rules

${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)={\mathrm{log}}_{3}\left(5\cdot 8\right)={\mathrm{log}}_{3}\left(40\right)$

This reduces our original expression to

${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)$

Then, using the quotient rule

${\mathrm{log}}_{3}\left(40\right)-{\mathrm{log}}_{3}\left(2\right)={\mathrm{log}}_{3}\left(\frac{40}{2}\right)={\mathrm{log}}_{3}\left(20\right)$

Try It

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

Solution

l o g (\frac{3 \cdot 5}{4 \cdot 6})

$\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right)$ ; can also be written

l o g (\frac{5}{8})

$\mathrm{log}\left(\frac{5}{8}\right)$ by reducing the fraction to lowest terms.

Example: Condensing Complex Logarithmic Expressions

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

Solution

We apply the power rule first:

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

Next we apply the product rule to the sum:

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

Finally, we apply the quotient rule to the difference:

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

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.

Solution

We apply the power rule first:

$2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)$

Next we apply the product rule to the sum:

$\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right)$

Finally, we apply the quotient rule to the difference:

$\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}\left({\left(3x+5\right)}^{{x}^{-1}}\right)}\right)$

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.

Solution

l o g (\frac{5 {(x - 1)}^{3} \sqrt{x}}{(7 x - 1)})

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

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

Solution

l o g \frac{x^{12} {(x + 5)}^{4}}{{(2 x + 3)}^{4}}

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

l o g {(\frac{x^{3} (x + 5)}{(2 x + 3)})}^{4}

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

Applications of Laws 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 it’s 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.
${pH} =-\log _{10}(a_{{\text{H}}^{+}})=\log _{10}\left({\frac {1}{a_{{\text{H}}^{+}}}}\right)$
For example, a solution with a hydrogen ion activity of $2×10^{-5}=\frac{1}{(2×10^5)}$ (at that level essentially the number of moles of hydrogen ions per liter of solution) has a pH of $\log_{10}(2×10^5)=5.3$

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

Example: Applying of the Laws 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?

Solution

Suppose C is the original concentration of hydrogen ions, and P is the original pH of the liquid. Then $P=-\mathrm{log}\left(C\right)$ . 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

$\text{pH}=-\mathrm{log}\left(2C\right)=-\left(\mathrm{log}\left(2\right)+\mathrm{log}\left(C\right)\right)=-\mathrm{log}\left(2\right)-\mathrm{log}\left(C\right)$

Since $P=-\mathrm{log}\left(C\right)$ , the new pH is

$\text{pH}=P-\mathrm{log}\left(2\right)\approx P - 0.301$

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?

Solution

Exponential and Logarithmic Equations and Models