Section 7.5 Algebraic Properties
Logarithms satisfy various algebraic properties directly resulting from corresponding exponential laws.
Theorem 7.27. Algebraic Properties of Logs.
Suppose \(a\neq 1\) is positive and suppose \(A\) and \(B\) are positive real numbers. Then,
\(\displaystyle \displaystyle \log_a(A\cdot B) = \log_a(A)+\log_a(B)\)
\(\displaystyle \displaystyle \log_a\left(\frac{A}{B}\right)= \log_a(A)-\log_a(B)\)
\(\displaystyle \displaystyle \log_a\left(A^n\right)= n\cdot \log_a(A)\)
In particular, for the natural logarithm with base \(e\text{,}\) we have:
\(\displaystyle \displaystyle \ln(A\cdot B) = \ln(A)+\ln(B)\)
\(\displaystyle \displaystyle \ln\left(\frac{A}{B}\right)= \ln(A)-\ln(B)\)
\(\displaystyle \displaystyle \ln\left(A^n\right)= n\cdot \ln(A)\)
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*}
Example 7.28. 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*}
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*}
Example 7.29. 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*}
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*}
Example 7.30. Using Properties of Logarithms.
Simplify the expression
\begin{equation*}
64^{\frac{1}{2}\log_8(2)}
\end{equation*}
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*}
