# 6.5 Logarithmic properties  (Page 5/10)

 Page 5 / 10

## Condensing complex logarithmic expressions

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

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:

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

## Rewriting as a single logarithm

Rewrite $\text{\hspace{0.17em}}2\mathrm{log}x-4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)\text{\hspace{0.17em}}$ as a single logarithm.

We apply the power rule first:

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

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

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

Finally, we apply the quotient rule to the difference:

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

Rewrite $\text{\hspace{0.17em}}\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x-1\right)+3\mathrm{log}\left(x-1\right)\text{\hspace{0.17em}}$ as a single logarithm.

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

Condense $\text{\hspace{0.17em}}4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right).$

$\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}};\text{\hspace{0.17em}}$ this answer could also be written $\text{\hspace{0.17em}}\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}.$

## Applying of the laws of logs

Recall that, in chemistry, $\text{\hspace{0.17em}}\text{pH}=-\mathrm{log}\left[{H}^{+}\right].\text{\hspace{0.17em}}$ If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Suppose $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is the original concentration of hydrogen ions, and $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is the original pH of the liquid. Then $\text{\hspace{0.17em}}P=–\mathrm{log}\left(C\right).\text{\hspace{0.17em}}$ If the concentration is doubled, the new concentration is $\text{\hspace{0.17em}}2C.$ 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 $\text{\hspace{0.17em}}P=–\mathrm{log}\left(C\right),$ 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.

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

The pH increases by about 0.301.

## Using the change-of-base formula for logarithms

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or $\text{\hspace{0.17em}}e,$ we use the change-of-base formula    to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.

To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms    .

Given any positive real numbers $\text{\hspace{0.17em}}M,b,$ and $\text{\hspace{0.17em}}n,$ where and $\text{\hspace{0.17em}}b\ne 1,$ we show

$\text{\hspace{0.17em}}{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$

Let $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}M.\text{\hspace{0.17em}}$ By taking the log base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ of both sides of the equation, we arrive at an exponential form, namely $\text{\hspace{0.17em}}{b}^{y}=M.\text{\hspace{0.17em}}$ It follows that

For example, to evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}36\text{\hspace{0.17em}}$ using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

## The change-of-base formula

The change-of-base formula    can be used to evaluate a logarithm with any base.

For any positive real numbers $\text{\hspace{0.17em}}M,b,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n,$ where and $\text{\hspace{0.17em}}b\ne 1,$

${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}.$

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

${\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}$

and

${\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}$

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