17.2: Laws of Logarithms
- Page ID
- 174385
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
\[ \log_b (1) = 0 \quad \text{and} \quad \log_b (b) = 1. \nonumber \]
- Proofs
- By definition of a logarithm, \( \log_b (1) \) is the exponent, \( p \), for which \( b^p = 1 \). Therefore, \( p = 0 \). As such,\[ \log_b(1) = 0. \nonumber \]Similarly, \( \log_b(b) \) is the exponent, \( q \), such that \( b^q = b \). Since \( b = b^1 \), this means that \( b^q = b^1 \). In other words, \( q = 1 \). Hence,\[ \log_b(b) = 1. \nonumber \]
\[ \log_b ( b^x ) = x \quad \text{and} \quad b^{\log_b (x)} = x, \text{ where }x > 0. \nonumber \]
- Proofs
- Consider \( \log_b(b^x) \). By the definition of a logarithm, this is the exponent, \( p \), such that \( b^p = b^x \). That is, \( x = p \). Thus, \( \log_b(b^x) = x \).
Now consider \( b^{\log_b(x)} \), where \( x > 0 \). Recall that \( \log_b(x) \) is the exponent, \( q \), such that \( b^q = x \). That is,\[ b^{\log_b(x)} = b^q = x. \nonumber \]
\[ \log_b (M) = \log_b (N) \text{ if and only if } M = N. \nonumber \]
- Proof
- From the definition of a logarithm, if we let \( \log_b(M) = m \) and \( \log_b(N) =n \), then \( b^m = M \) and \( b^n = N \).
(\( \implies \)): If \( \log_b(M) = \log_b(N) \), then \( m = n \). However, if \( m = n \), then \( b^m = b^n \). Therefore, \( M = N \).
(\( \impliedby \)): If \( M = N \), then \( b^m = b^n \). However, by laws of exponents, this means \( m = n \). Hence, \( \log_b (M) = \log_b(N) \).
Given any real number \(x\), and positive real numbers \(M\) and \(N\),\[ \log_b(MN) = \log_b (M) + \log_b (N). \nonumber \]
- Proof
- 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\[ \begin{array}{rclcl} \log_b(MN) & = & \log_b(b^m b^n) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b(b^{m + n}) & \quad & \left( \text{Laws of Exponents: Product Rule} \right) \\[6pt] & = & m + n & \quad & \left( \text{Laws of Logarithms: Inverse Law} \right) \\[6pt] & = & \log_b(M) + \log_b(N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]
Given any real number \(x\), and positive real numbers \(M\) and \(N\),\[ \log_b \left( \dfrac{M}{N} \right) = \log_b ( M )− \log_b ( N ). \nonumber \]
- Proof
- 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\[ \begin{array}{rclcl} \log_b \left( \dfrac{M}{N} \right) & = & \log_b \left( \dfrac{b^m}{b^n} \right) & \quad & \left( \text{substituting} \right) \\[6pt] & = & \log_b \left( b^{m - n} \right) & \quad & \left( \text{Laws of Exponents: Quotient Law} \right) \\[6pt] & = & m - n & \quad & \left( \text{Laws of Exponents: Inverse Law} \right) \\[6pt] & = & \log_b (M) - \log_b (N) & \quad & \left( \text{substituting} \right) \\[6pt] \end{array} \nonumber \]
\[ \log_b ( M^p ) = p \log_b (M). \nonumber \]
- Proof
- Let \( \log_b (M) = m \).\[ \begin{array}{rrclcl}
\implies & b^m & = & M & \quad & \left( \text{definition of the logarithm} \right) \\[6pt]
\implies & \left( b^m \right)^p & = & M^p & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & b^{mp} & = & M^p & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \log_b \left(b^{mp}\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{One-to-One Law of Logarithms} \right) \\[6pt]
\implies & \log_b \left(b^m \cdot b^p\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \log_b \left(b^m \right) \cdot \log_b \left( b^p\right) & = & \log_b \left(M^p\right) & \quad & \left( \text{Product Law of Logarithms} \right) \\[6pt]
\implies & m \cdot p & = & \log_b \left(M^p\right) & \quad & \left( \text{Base Laws of Logarithms} \right) \\[6pt]
\implies & \log_b(M) \cdot p & = & \log_b \left(M^p\right) & \quad & \left( \text{our original substitution }\log_b(M) = m \right) \\[6pt]
\implies & p \log_b(M) & = & \log_b \left(M^p\right) & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt]
\end{array} \nonumber \]
Given any positive real numbers \(M\), \(b\), and \(n\), where \(n \neq 1\), and \(b \neq 1\),\[ \log_b (M) = \dfrac{\log_n (M)}{\log_n (b)}. \nonumber \]
- Proof
- Let \(y= \log_b (M)\). By the definition of a logarithm, \(b^y =M\). It follows that\[ \begin{array}{rrclcl}
& \log_n ( b^y ) & = & \log_n (M) & \quad & \left( \text{One-to-One Law of Logarithms} \right) \\[6pt]
\implies & y \log_n (b) & = & \log_n (M) & \quad & \left( \text{Power Law of Logarithms} \right) \\[6pt]
\implies & y & = & \dfrac{\log_n (M)}{\log_n (b)} & \quad & \left( \text{dividing both sides by }\log_n (b) \right) \\[6pt]
\implies & \log_b (M) & = & \dfrac{\log_n (M)}{\log_n (b)} & \quad & \left( \text{substituting} \right) \\[6pt]
\end{array} \nonumber \]