Algebraic Properties

Section 8.5 Algebraic Properties

Logarithms satisfy various algebraic properties directly resulting from corresponding exponential laws (see Section 3.6).

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Theorem 8.5.1. Algebraic Properties of Logs.

Suppose \(a\neq 1\) is positive and suppose \(A\) and \(B\) are positive real numbers. Then,

\(\displaystyle \displaystyle \log_a(A\cdot B) = \log_a(A)+\log_a(B)\)

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\(\displaystyle \displaystyle \log_a\left(\frac{A}{B}\right)= \log_a(A)-\log_a(B)\)

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\(\displaystyle \displaystyle \log_a\left(A^n\right)= n\cdot \log_a(A)\)

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In particular, for the natural logarithm with base \(e\text{,}\) we have:

\(\displaystyle \displaystyle \ln(A\cdot B) = \ln(A)+\ln(B)\)

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\(\displaystyle \displaystyle \ln\left(\frac{A}{B}\right)= \ln(A)-\ln(B)\)

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\(\displaystyle \displaystyle \ln\left(A^n\right)= n\cdot \ln(A)\)

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Proof.

Let’s prove the first statement for the natural logarithm, leaving the rest as exercises. First, let

\begin{equation*} x = \ln(A) \quad \text{and} \quad y = \ln(B). \end{equation*}

In exponential form, we have

\begin{equation*} e^x = A \quad \text{and} \quad e^y = B. \end{equation*}

Multiplying, and using the appropriate Law of Exponents,

\begin{equation*} A\cdot B = e^x \cdot e^y = e^{\boxed{x+y}} \end{equation*}

which in logarithmic form says

\begin{equation*} \boxed{x+y} = \ln(A\cdot B). \end{equation*}

Rewriting \(x\) and \(y\)

\begin{equation*} \ln(A)+\ln(B) = \ln(A\cdot B). \end{equation*}

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Example 8.5.2. Using Properties of Logarithms.

Given that \(\ln(x) = 2\text{,}\) \(\ln(y)=5\text{,}\) and \(\ln(z)=4\text{,}\) use the properties of logarithms to find the value of

\begin{equation*} \ln\left(\frac{x^3}{y \sqrt{z}}\right). \end{equation*}

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Solution.

Expanding,

\begin{align*} \ln\left(\frac{x^3 y}{\sqrt{z}}\right) &= \ln(x^3) - \ln(y \sqrt{z}) \\ &= \ln(x^3) - \ln(y \sqrt{z}) \\ &= \ln(x^3) - \left[\ln(y)+\ln(\sqrt{z})\right] \\ &= 3\ln(x) - \ln(y) -\ln(z^{1/2}) \\ &= 3\ln(x) - \ln(y) -\frac{1}{2} \ln(z) \\ &= 3\cdot 2 - 5 -\frac{1}{2}\cdot 4 = \boxed{-1}. \end{align*}

In exponential form, this says,

\begin{equation*} e^{\boxed{-1}} = \frac{x^3 y}{\sqrt{z}} \end{equation*}

or that

\begin{equation*} \frac{x^3 y}{\sqrt{z}} = \frac{1}{e}. \end{equation*}

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Example 8.5.3. Using Properties of Logarithms.

Express as a single logarithm.

\begin{equation*} 4 \log_3(x+1)- 2 \log_3(x) - \log_3(x). \end{equation*}

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Solution.

Applying the properties of logarithms in reverse now,

\begin{align*} 4 \log_3(x+1)- 2 \log_3(x) - \log_3(x) &= \log_3((x+1)^3)- \log_3(x^2) - \log_3(x)\\ &= \log_3\left(\frac{(x+1)^3}{x^2}\right) - \log_3(x)\\ &= \log_3\left(\frac{(x+1)^3}{x^2\cdot x}\right)\\ &= \log_3\left(\frac{(x+1)^3}{x^3}\right) \end{align*}

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Example 8.5.4. Using Properties of Logarithms.

Simplify the expression

\begin{equation*} 64^{\frac{1}{2}\log_8(2)} \end{equation*}

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Solution.

We have

\begin{align*} 64^{\frac{1}{2}\log_8(2)} &= \left(8^2\right)^{\log_8(2^{1/2})}\\ &= 8^{2\log_8(\sqrt{2})}\\ &= 8^{\log_8(2)}\\ &= 2 \end{align*}

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