## 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.

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 ${\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)$ 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 9

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

Solution

### Example 10: 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 11: 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 10

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

### Try It 11

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

Solution

### Example 12: 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 12

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

Solution