4.5 Logarithmic properties (Page 5/10)

Precalculus

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Condensing complex logarithmic expressions

Condense $\log_{2} (x^{2}) + \frac{1}{2} \log_{2} (x - 1) - 3 \log_{2} ({(x + 3)}^{2}) .$

We apply the power rule first:

\log_{2} (x^{2}) + \frac{1}{2} \log_{2} (x - 1) - 3 \log_{2} ({(x + 3)}^{2}) = \log_{2} (x^{2}) + \log_{2} (\sqrt{x - 1}) - \log_{2} ({(x + 3)}^{6})

Next we apply the product rule to the sum:

\log_{2} (x^{2}) + \log_{2} (\sqrt{x - 1}) - \log_{2} ({(x + 3)}^{6}) = \log_{2} (x^{2} \sqrt{x - 1}) - \log_{2} ({(x + 3)}^{6})

Finally, we apply the quotient rule to the difference:

\log_{2} (x^{2} \sqrt{x - 1}) - \log_{2} ({(x + 3)}^{6}) = \log_{2} \frac{x^{2} \sqrt{x - 1}}{{(x + 3)}^{6}}

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Rewriting as a single logarithm

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

We apply the power rule first:

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

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

\log (x^{2}) - \log {(x + 5)}^{4} + \log ({(3 x + 5)}^{x^{- 1}})

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

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

Finally, we apply the quotient rule to the difference:

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

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

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

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

$\log \frac{x^{12} {(x + 5)}^{4}}{{(2 x + 3)}^{4}};$ this answer could also be written $\log {(\frac{x^{3} (x + 5)}{(2 x + 3)})}^{4} .$

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Applying of the laws of logs

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

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

pH = - \log (2 C)

Using the product rule of logs

pH = - \log (2 C) = - (\log (2) + \log (C)) = - \log (2) - \log (C)

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

pH = P - \log (2) \approx P - 0.301

When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

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

The pH increases by about 0.301.

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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 $e,$ 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 $M, b,$ and $n,$ where $n \neq 1$ and $b \neq 1,$ we show

\log_{b} M = \frac{\log_{n} M}{\log_{n} b}

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

\begin{array}{l} \log_{n} (b^{y}) & = \log_{n} M & Apply the one-to-one property . \\ y \log_{n} b & = \log_{n} M & Apply the power rule for logarithms . \\ y & = \frac{\log_{n} M}{\log_{n} b} & Isolate y . \\ \log_{b} M & = \frac{\log_{n} M}{\log_{n} b} & Substitute for y . \end{array}

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

\begin{array}{l} \log_{5} 36 & = \frac{\log (36)}{\log (5)} & Apply the change of base formula using base 10 . \\ \approx 2.2266 & Use a calculator to evaluate to 4 decimal places . \end{array}

A General Note

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 $M, b,$ and $n,$ where $n \neq 1$ and $b \neq 1,$

\log_{b} M = \frac{\log_{n} M}{\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.

\log_{b} M = \frac{\ln M}{\ln b}

and

\log_{b} M = \frac{\log M}{\log b}

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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