MA001: College Algebra

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:

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

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:

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

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

Solution

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

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

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

$\ln (x^4y) − \ln (7) = \ln (x^4) + \ln (y) − \ln (7)$

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

$\ln (x^4) + \ln (y) − \ln (7) = 4 \ln (x) + \ln (y) − \ln (7)$

TRY IT #6

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

EXAMPLE 7

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

Expand $log(\sqrt{x})$.

Solution

$\begin{aligned} \log (\sqrt{x}) &=\log x^{\left(\frac{1}{2}\right)} \\ &=\frac{1}{2} \log x \end{aligned}$

TRY IT #7

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

Q&A

Can we expand $\ln (x^2+y^2)$?

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 $log_6 \left(\frac{64x^3(4x+1)}{(2x−1)}\right)$.

Solution

We can expand by applying the Product and Quotient Rules.

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

TRY IT #8

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

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 $log_3(5)+log_3(8)−log_3(2)$ as a single logarithm.

Solution

Using the product and quotient rules

$log_3(5)+log_3(8)=log_3(5⋅8)=log_3(40)$

This reduces our original expression to

$log_3(40)−log_3(2)$

Then, using the quotient rule

$log_3(40)−log_3(2)=log_3 \left(\frac{40}{2}\right)=log_3(20)$

TRY IT #9

Condense $log3−log4+log5−log6$.

EXAMPLE 10

Condensing Complex Logarithmic Expressions

Condense $log_2(x^2)+\frac{1}{2}log_2(x−1)−3log_2((x+3)^2)$.

Solution

We apply the power rule first:

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

Next we apply the product rule to the sum:

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

Finally, we apply the quotient rule to the difference:

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

TRY IT #10

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

EXAMPLE 11

Rewriting as a Single Logarithm

Rewrite $2logx−4log(x+5)+ \frac{1}{x}log(3x+5)$ as a single logarithm.

Solution

We apply the power rule first:

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

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

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

Finally, we apply the quotient rule to the difference:

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

TRY IT #11

Condense $4(3 \log (x)+\log (x+5)-\log (2 x+3))$

EXAMPLE 12

Applying of the Laws of Logs

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

Solution

Suppose $C$ is the original concentration of hydrogen ions, and $P$ is the original $\mathrm{pH}$ of the liquid. Then $P=-\log (C)$. If the concentration is doubled, the new concentration is $2 C$. Then the $\mathrm{pH}$ of the new liquid is

$\mathrm{pH}=-\log (2 C)$

Using the product rule of logs

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

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

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

When the concentration of hydrogen ions is doubled, the $\mathrm{pH}$ decreases by about $0.301$.

TRY IT #12

How does the $\mathrm{pH}$ change when the concentration of positive hydrogen ions is decreased by half?

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-5-logarithmic-properties
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