Logarithms

\(\displaystyle{ {\log\mathopen{}\left(\frac{x^{8}z^{8}}{y^{6}}\right)} = }\)
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\(\displaystyle{ {\log\mathopen{}\left(\frac{x^{6}}{y^{8}z^{8}}\right)} = }\)
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\(\displaystyle{ {\log\mathopen{}\left(\left(\frac{x^{8}}{y^{8}z^{8}}\right)^{9}\right)} = }\)
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Hint.

Do you have a logarithm of a fraction? If so, what is the \(\color{red}{numerator}\text{?}\) And the \(\color{blue}{denominator}\text{?}\) Rewrite your expression using that ` log(frac{color{red}{a}}{color{blue}{b}}) = log(color{red}{a}) - log(color{blue}{b}) `.
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Or do you have a logarithm of a power? If so, what is the \(\color{red}{base}\text{?}\) And the \(\color{blue}{exponent}\text{?}\) Rewrite your expression using that `log(color{red}{b}^color{blue}{a}) = color{blue}{a}log (color{red}{b})`.
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Do you have a logarithm of a product? What are the \(\color{green}{factors}\text{?}\) If so, use that ` log(color{green}{cd}) = log (color{green}{c}) + log(color{green}{d}) `.
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If you have a negative logarithm like ` color{brown}{-}log(cd) `, make sure to use parentheses:
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` color{brown}{-}log(cd) rightarrow color{brown}{-} (log(c)+log(d)) rightarrow color{brown}{-}log(c)color{brown}{-}log(d)`

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Answer 1.

\(8\log\mathopen{}\left(x\right)+8\log\mathopen{}\left(z\right)-6\log\mathopen{}\left(y\right)\)

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Answer 2.

\(6\log\mathopen{}\left(x\right)-8\log\mathopen{}\left(y\right)-8\log\mathopen{}\left(z\right)\)

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Answer 3.

\(72\log\mathopen{}\left(x\right)-72\log\mathopen{}\left(y\right)-72\log\mathopen{}\left(z\right)\)

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Solution.

The expression is a logarithm of a fraction.
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\(\begin{aligned} \amp \log\left(\dfrac{x^{8} z^{8}}{y^{6}}\right) \amp \text{log of a } \textbf{fraction}\\ \longrightarrow\quad\amp \log\left(\dfrac{\color{red}{x^{8} z^{8}}}{\color{blue}{y^{6}}}\right) \amp \text{identify the} \;\color{red}{numerator}\;\text{ and the }\;\color{blue}{denominator} \\ \amp \amp \text{ use that } \log\left(\frac{\color{red}{a}}{\color{blue}{b}}\right) = \log(\color{red}{a}) - \log(\color{blue}{b})\\ \\ \longrightarrow\quad \amp \log( \color{red}{x^{8} z^{8}} )-\log(\color{blue}{y^{6}}) \amp \text{the first log is a log of a } \textbf{product}\\ \\ \longrightarrow\quad \amp \log( \color{red}{x^{8}}\color{blue}{z^{8}} )-\log(y^{6}) \amp \text{there are two factors; identify } \color{red}{factor\; 1\;} \text{and} \color{blue}{\; factor \; 2}\\ \amp \amp \text{ use that } \log(\color{red}{c}\color{blue}{d}) = \log (\color{red}{c}) + \log(\color{blue}{d}) \\ \\ \longrightarrow\quad \amp \log( \color{red}{x^{8}})+\log(\color{blue}{z^{8}} )-\log(y^{6}) \amp \text{three logs of } \textbf{powers} \\ \\ \longrightarrow\quad \amp \log( \color{red}{x}^{\color{blue}8})+\log(\color{red}{z}^{\color{blue}8} )-\log(\color{red}{y}^{\color{blue}6}) \amp \text{identify } \color{red}{bases\;} \text{and} \color{blue}{\; exponents} \\ \amp \amp \text{ use that } \log(\color{red}{b}^{\color{blue}a}) = \color{blue}{a}\log (\color{red}{b})\\ \\ \longrightarrow\quad \amp \color{blue}{8}\log(\color{red}{x})+\color{blue}{8}\log(\color{red}{z})-\color{blue}{6}\log(\color{red}{y}) \amp \\ \end{aligned}\)
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The expression is a logarithm of a fraction.
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\(\begin{aligned} \amp \log\left(\dfrac{x^6}{y^8z^8}\right) \amp \text{log of a }\textbf{fraction}\\ \\ \longrightarrow\quad\amp \log\left(\dfrac{\color{red}{x^6}}{\color{blue}{y^8z^8}}\right) \amp \text{identify the} \;\color{red}{numerator}\;\text{ and the }\;\color{blue}{denominator} \\ \amp \amp \text{ use that } \log\left(\frac{\color{red}{a}}{\color{blue}{b}}\right) = \log(\color{red}{a}) - \log(\color{blue}{b})\\ \\ \longrightarrow\quad \amp \log( \color{red}{x^6} )-\log(\color{blue}{y^8z^8}) \amp \text{the second log is a log of a }\textbf{product}\\ \\ \longrightarrow\quad \amp \log(x^6)-\log(\color{red}{y^8}\color{blue}{z^8}) \amp \text{there are two factors; identify } \color{red}{factor\; 1\;} \text{and} \color{blue}{\; factor \; 2}\\ \amp \amp \text{ use that } \log(\color{red}{c}\color{blue}{d}) = \log (\color{red}{c}) + \log(\color{blue}{d}) \\ \amp \amp \text{ don't forget to use parentheses}\\ \\ \longrightarrow\quad \amp \log( x^6)- (\log(\color{red}{y^8} )+ \log(\color{blue}{z^8})) \amp \text{three logs of }\textbf{powers} \\ \\ \longrightarrow\quad \amp \log( \color{red}{x}^{\color{blue} {6}})-\log(\color{red}{y}^{\color{blue}{8}} )-\log(\color{red}{z}^{\color{blue}{8}}) \amp \text{distribute the negative sign} \\ \amp \amp \text{identify } \color{red}{bases\;} \text{and} \color{blue}{\; exponents} \\ \amp \amp \text{ use that } \log(\color{red}{b}^{\color{blue}{a}}) = \color{blue}{a}\log (\color{red}{b})\\ \\ \longrightarrow\quad \amp \color{blue}{6}\log(\color{red}{x})-\color{blue}{8}\log(\color{red}{y})-\color{blue}{8}\log(\color{red}{z}) \amp \\ \end{aligned}\)
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The expression is a logarithm of a power.
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\(\begin{aligned} \amp \log\left( \left(\dfrac{x^8}{y^8 z^8}\right)^{9} \right) \amp \text{log of a }\textbf{power}\\ \\ \longrightarrow\quad \amp \log\left( \left(\color{red}{\dfrac{x^8}{y^8 z^8}}\right)^{\color{blue}{9}} \right) \amp \text{identify the} \;\color{red}{base}\;\text{ and the }\;\color{blue}{exponent} \\ \amp \amp \text{ use that } \log(\color{red}{b}^{\color{blue}{a}}) = \color{blue}{a}\log (\color{red}{b})\\ \\ \longrightarrow\quad \amp \color{blue}{9} \log\left(\color{red}{\dfrac{x^8}{y^8 z^8}}\right) \amp \text{the log expression is a log of a }\textbf{fraction}\\ \\ \longrightarrow\quad\amp 9 \log\left(\dfrac{\color{red}{x^8}}{\color{blue}{y^8 z^8}}\right) \amp \text{identify the} \;\color{red}{numerator}\;\text{ and the }\;\color{blue}{denominator} \\ \amp \amp \text{ use that } \log\left(\frac{\color{red}{a}}{\color{blue}{b}}\right) = \log(\color{red}{a}) - \log(\color{blue}{b})\\ \amp \amp \text{ don't forget to use parentheses}\\ \\ \longrightarrow\quad \amp 9\left( \log(\color{red}{x^8})-\log(\color{blue}{y^8 z^8})\right) \amp \text{the second log is a log of a }\textbf{product}\\ \\ \longrightarrow\quad \amp 9\left( \log(x^8)-\log(\color{red}{y^8} \color{blue}{z^8})\right) \amp \text{there are two factors; identify } \color{red}{factor\; 1\;} \text{and} \color{blue}{\; factor \; 2}\\ \amp \amp \text{ use that } \log(\color{red}{c}\color{blue}{d}) = \log (\color{red}{c}) + \log(\color{blue}{d}) \\ \amp \amp \text{ don't forget to use parentheses}\\ \\ \longrightarrow\quad \amp 9\left( \log(x^8)-\left(\log(\color{red}{y^8})+\log(\color{blue}{z^8})\right)\right) \amp \text{three logs of }\textbf{powers} \\ \\ \longrightarrow\quad \amp 9\left( \log(\color{red}{x}^{\color{blue}{8}})-\log(\color{red}{y}^{\color{blue}{8}})-\log(\color{red}{z}^{\color{blue}{8}})\right)\amp \text{distribute the negative sign} \\ \amp \amp \text{identify } \color{red}{bases\;} \text{and} \color{blue}{\; exponents} \\ \amp \amp \text{ use that } \log(\color{red}{b}^{\color{blue}{a}}) = \color{blue}{a}\log (\color{red}{b})\\ \\ \longrightarrow\quad \amp 9\left(\color{blue}{8}\log(\color{red}{x})-\color{blue}{8}\log(\color{red}{y})-\color{blue}{8}\log(\color{red}{z})\right) \amp \text{multiply each term within parentheses by 9}\\ \\ \longrightarrow\quad \amp 72\log(x)-72\log(y)-72\log(z)\amp \\ \end{aligned}\)
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8.

Use the properties of logarithms to expand the following logarithm as much as possible. Rewrite as a sum, difference, or product of logs.

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\(\log{\left( x^9y^6\sqrt[3]{x^3y^7}\right)}\)

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Answer.

\(10\log\mathopen{}\left(x\right)+8.33333\log\mathopen{}\left(y\right)\)

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9.

Condense the following expression to a single logarithm using the properties of logarithms.

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\(\ln{\left( 3x^6\right)}- \ln{\left( 5x^5\right)}\)

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Answer.

\(\ln\mathopen{}\left(0.6x^{1}\right)\)

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10.

Use the change of base formula to convert \(\log_{3}(12)\) to a ratio of logs base \(e\)

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\(\displaystyle \displaystyle \frac{\log_{12}(12)}{\log_{12}(3)}\)
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\(\displaystyle \displaystyle \frac{\log(12)}{\log(3)}\)
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\(\displaystyle \displaystyle \frac{\ln(12)}{\ln(3)}\)
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\(\displaystyle \displaystyle \frac{\ln(3)}{\ln(12)}\)
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None of the above
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Use the change of base formula to convert \(\log_{8}(4)\) to a ratio of logs base \(2\)

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\(\displaystyle \displaystyle \frac{\log_{2}(8)}{\log_{2}(4)}\)
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\(\displaystyle \displaystyle \frac{\log_{4}(2)}{\log_{4}(8)}\)
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\(\displaystyle \displaystyle \frac{\log_{2}(4)}{\log_{2}(8)}\)
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\(\displaystyle \displaystyle \frac{\log(4)}{\log(8)}\)
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None of the above
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Hint.

Change-of-Base Formula

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

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For any positive real numbers \(M, b\text{,}\) and \(n\text{,}\) where \(n\neq 1\) and \(b\neq 1\text{,}\)

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\(\displaystyle{ \log_bM = \frac{\log_nM}{\log_nb}}\)

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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.

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\(\displaystyle{ \log_bM = \frac{\ln M}{\ln b}}\)

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and

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\(\displaystyle{ \log_bM = \frac{\log M}{\log b}}\)

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11.

Suppose \(\log_{8}(4)=a\) and \(\log_{8}(5)=b\text{.}\) Use the change of base formula along with properties of logarithms to rewrite the following in terms of \(a\) and \(b\text{.}\)

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\(\log_{5}(8)=\)
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\(\log_{5}\left(\frac{4}{5}\right)=\)
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Answer 1.

\(\frac{1}{b}\)

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Answer 2.

\(\frac{a}{b}-1\)

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12.

Rewrite the following logarithms in expanded form by applying the properties of logarithms.

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\({\log\mathopen{}\left(x^{y}\right)} =\)
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\({\log\mathopen{}\left(\frac{x}{y}\right)} =\)
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\({\log\mathopen{}\left(xy\right)} =\)
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Answer 1.

\(y\log\mathopen{}\left(x\right)\)

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Answer 2.

\(\log\mathopen{}\left(x\right)-\log\mathopen{}\left(y\right)\)

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Answer 3.

\(\log\mathopen{}\left(x\right)+\log\mathopen{}\left(y\right)\)

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Solution.

` log(xy) = log(x)+log(y)`

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` log(frac{x}{y}) =log(x)-log(y)`

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` log(x^y) = ylog(x)`

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13.

Rewrite \({\frac{7}{2}}\log_{8} x+3\log_{8} y-\log_{8} z-{\frac{1}{6}}\log_{8} w\) as a single logarithm.

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\(\displaystyle \displaystyle\log_{8} \left(\frac{ {x^8} {y^8}}{\sqrt[3] {z}\sqrt[5] {w^4}}\right)\)
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\(\displaystyle \displaystyle\log_{8} \left(\frac{\sqrt {x^3}\sqrt[8] {y}}{\sqrt[3] {z}\sqrt[3] {w}}\right)\)
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\(\displaystyle \displaystyle\log_{8} \left(\frac{\sqrt {x^7} {y^3}}{ {z}\sqrt[6] {w}}\right)\)
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\(\displaystyle \displaystyle\log_{8} \left(\frac{\sqrt[4] {x^7}\sqrt[4] {y}}{ {z} {w}}\right)\)
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Answer.

\(\text{Choice 3}\)

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14.

Expand \(\displaystyle \log\left(\frac{a^2}{b^{-3}c^4}\right)\) to rewrite as a sum, difference, or product of logarithms. Choose all correct answers. There may be more than one way to rewrite the expression.

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