# 6.5 Logarithmic properties  (Page 4/10)

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

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:

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

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.

## Expanding logarithms using product, quotient, and power rules

Rewrite $\text{\hspace{0.17em}}\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)\text{\hspace{0.17em}}$ as a sum or difference of logs.

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:

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

Expand $\text{\hspace{0.17em}}\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right).$

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

## Using the power rule for logarithms to simplify the logarithm of a radical expression

Expand $\text{\hspace{0.17em}}\mathrm{log}\left(\sqrt{x}\right).$

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

Expand $\text{\hspace{0.17em}}\mathrm{ln}\left(\sqrt[3]{{x}^{2}}\right).$

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

Can we expand $\text{\hspace{0.17em}}\mathrm{ln}\left({x}^{2}+{y}^{2}\right)?$

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.

## Expanding complex logarithmic expressions

Expand $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x-1\right)}\right).$

We can expand by applying the Product and Quotient Rules.

Expand $\text{\hspace{0.17em}}\mathrm{ln}\left(\frac{\sqrt{\left(x-1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right).$

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

## 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.

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.

## Using the product and quotient rules to combine logarithms

Write $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)\text{\hspace{0.17em}}$ as a single logarithm.

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

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

Condense $\text{\hspace{0.17em}}\mathrm{log}3-\mathrm{log}4+\mathrm{log}5-\mathrm{log}6.$

$\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right);\text{\hspace{0.17em}}$ can also be written $\text{\hspace{0.17em}}\mathrm{log}\left(\frac{5}{8}\right)\text{\hspace{0.17em}}$ by reducing the fraction to lowest terms.

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