## Expand 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:

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

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:

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

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 $\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)$ 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 6

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

Solution

### Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression

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

### Solution

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

### Try It 7

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

Solution

### 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.

### Example 8: 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.

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

### 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