6.5 Logarithmic properties (Page 2/10)

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\log_{b} (M N) = \log_{b} (M) + \log_{b} (N) .

Let $m = \log_{b} M$ and $n = \log_{b} N .$ In exponential form, these equations are $b^{m} = M$ and $b^{n} = N .$ It follows that

\begin{array}{l} \log_{b} (M N) & = \log_{b} (b^{m} b^{n}) & Substitute for M and N . \\ = \log_{b} (b^{m + n}) & Apply the product rule for exponents . \\ = m + n & Apply the inverse property of logs . \\ = \log_{b} (M) + \log_{b} (N) & Substitute for m and n . \end{array}

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider $\log_{b} (w x y z) .$ Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

\log_{b} (w x y z) = \log_{b} w + \log_{b} x + \log_{b} y + \log_{b} z

A General Note

The product rule for logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

\log_{b} (M N) = \log_{b} (M) + \log_{b} (N) for b > 0

How To

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

Factor the argument completely, expressing each whole number factor as a product of primes.
Write the equivalent expression by summing the logarithms of each factor.

Using the product rule for logarithms

Expand $\log_{3} (30 x (3 x + 4)) .$

We begin by factoring the argument completely, expressing $30$ as a product of primes.

\log_{3} (30 x (3 x + 4)) = \log_{3} (2 \cdot 3 \cdot 5 \cdot x \cdot (3 x + 4))

Next we write the equivalent equation by summing the logarithms of each factor.

\log_{3} (30 x (3 x + 4)) = \log_{3} (2) + \log_{3} (3) + \log_{3} (5) + \log_{3} (x) + \log_{3} (3 x + 4)

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Expand $\log_{b} (8 k) .$

$\log_{b} 2 + \log_{b} 2 + \log_{b} 2 + \log_{b} k = 3 \log_{b} 2 + \log_{b} k$

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Using the quotient rule for logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: $x^{\frac{a}{b}} = x^{a - b} .$ The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number $x$ and positive real numbers $M,$ $N,$ and $b,$ where $b \neq 1,$ we will show

\log_{b} (\frac{M}{N}) = \log_{b} (M) - \log_{b} (N) .

Let $m = \log_{b} M$ and $n = \log_{b} N .$ In exponential form, these equations are $b^{m} = M$ and $b^{n} = N .$ It follows that

\begin{array}{l} \log_{b} (\frac{M}{N}) & = \log_{b} (\frac{b^{m}}{b^{n}}) & Substitute for M and N . \\ = \log_{b} (b^{m - n}) & Apply the quotient rule for exponents . \\ = m - n & Apply the inverse property of logs . \\ = \log_{b} (M) - \log_{b} (N) & Substitute for m and n . \end{array}

For example, to expand $\log (\frac{2 x^{2} + 6 x}{3 x + 9}),$ we must first express the quotient in lowest terms. Factoring and canceling we get,

\begin{array}{l} \log (\frac{2 x^{2} + 6 x}{3 x + 9}) = \log (\frac{2 x (x + 3)}{3 (x + 3)}) & Factor the numerator and denominator . \\ = \log (\frac{2 x}{3}) & Cancel the common factors . \end{array}

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

\begin{array}{l} \log (\frac{2 x}{3}) = \log (2 x) - \log (3) \\ = \log (2) + \log (x) - \log (3) \end{array}

A General Note

The quotient rule for logarithms

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

\log_{b} (\frac{M}{N}) = \log_{b} M - \log_{b} N

How To

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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