Laws of Logarithms II
Laws of Logarithms II
Logarithmic Law:
\[\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\]\begin{align*} \text{Let } {\log}_{a}(x) = m & \quad \implies \quad x = a^{m} \ldots (1)\qquad (x > 0 )\\ \text{and }{\log}_{a}(y) = n & \quad \implies \quad x = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \times (2): \quad x \times y &= a^{m} \times a^{n}\\ \therefore xy &= a^{m + n} \end{align*}
Now we change from the exponential form back to logarithmic form:
\begin{align*} \log_{a}{xy} &= m + n \\ \text{But } m = {\log}_{a}(x) & \text{ and } n = {\log}_{a}(y) \\ \therefore \log_{a}{xy} &= {\log}_{a}(x) +{\log}_{a}(y) \end{align*}
In words: the logarithm of a product is equal to the sum of the logarithms of the factors.
Example
Question
Simplify: \(\log{5} + \log{2} - \log{30}\)
Use the logarithmic law to simplify the expression
We combine the first two terms since the product of \(\text{5}\) and \(\text{2}\) is equal to \(\text{10}\), which is always useful when simplifying logarithms.
\begin{align*} \log{5} + \log{2} - \log{30} &= ( \log{5} + \log{2} ) - \log{30} \\ &= \log{( 5 \times 2)} - \log{30} \\ &= \log{10} - \log{30} \\ &= 1 - \log{30} \end{align*}We expand the last term to simplify the expression further:
\begin{align*} &= 1 - \log{( 3 \times 10)} \\ &= 1 - ( \log{3} + \log{10} ) \\ &= 1 - ( \log{3} + 1 ) \\ &= 1 - \log{3} - 1 \\ &= - \log{3} \end{align*}Write the final answer
\(\log{5} + \log{2} - \log{30} = - \log{3}\)
Logarithmic law:
\[\log_{a}{\cfrac{x}{y}} = \log_{a}{x} - \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\]\begin{align*} \text{Let } {\log}_{a}(x) = m & \quad \implies \quad x = a^{m} \ldots (1) \qquad (x > 0)\\ \text{and }{\log}_{a}(y) = n & \quad \implies \quad y = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \div (2): \quad \cfrac{x}{y} &= \cfrac{a^{m}}{a^{n}} \\ \therefore \cfrac{x}{y} &= a^{m - n} \end{align*}
Now we change from the exponential form back to logarithmic form:
\begin{align*} \log_{a}{\cfrac{x}{y}} &= m - n \\ \text{But } m = {\log}_{a}(x) & \text{ and } n = {\log}_{a}(y) \\ \therefore \log_{a}{\cfrac{x}{y}} &= {\log}_{a}(x) - {\log}_{a}(y) \end{align*}
In words: the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator.
Example
Question
Simplify: \(\log{40} - \log{4} + \log_{5}{\cfrac{8}{5}}\)
Use the logarithmic law to simplify the expression
We combine the first two terms since both terms have the same base and the quotient of \(\text{40}\) and \(\text{4}\) is equal to \(\text{10}\):
\begin{align*} \log{40} - \log{4} + \log_{5}{\cfrac{8}{5}} &= (\log{40} - \log{4} ) + \log_{5}{\cfrac{8}{5}} \\ &= (\log{\cfrac{40}{4}} ) + \log_{5}{\cfrac{8}{5}} \\ &= \log{10} + \log_{5}{\cfrac{8}{5}} \\ &= 1 + \log_{5}{\cfrac{8}{5}} \end{align*}We expand the last term to simplify the expression further:
\begin{align*} &= 1 + ( \log_{5}{8} - \log_{5}{5} ) \\ &= 1 + \log_{5}{8} - 1 \\ &= \log_{5}{8} \end{align*}Write the final answer
\(\log{40} - \log{4} + \log_{5}{\cfrac{8}{5}} = \log_{5}{8}\)
Useful summary:
-
\(\log1=0\)
-
\(\log10=1\)
-
\(\log100=2\)
-
\(\log1000=3\)
-
\(\log{\cfrac{1}{10}}=-1\)
-
\(\log{\text{0.1}}=-1\)
-
\(\log{\text{0.01}}=-2\)
-
\(\log{\text{0.001}}=-3\)
This lesson is part of:
Functions III