Rules for logarithms

Exponential functions and logarithms: Logarithmic functions

Rules for logarithms

We have already studied some important rules and definitions regarding the logarithm. You will now learn three additional rules, which come in handy when solving logarithmic equations. Along with the same restrictions for $\blue{a}$ and $\green{b}$ , it also holds that $\purple{c}$ should be positive.

$\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)+\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\green{b}\cdot \purple{c}\right)\end{array}$

Example

$\begin{array}{lccr}\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)&=&\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)\end{array}$

Note that $\begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right)}&=&\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\cdot \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}\\&=&\green{b}\cdot \purple{c}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}\cdot \purple{c}\right)}\end{array}$ From this, we conclude that $\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right) = \log_\blue{a}\left(\green{b}\cdot \purple{c}\right).$

Example

$\begin{array}{rcl}\blue{3}^{\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)}&=&\blue{3}^{\log_\blue{3}\left(\green{6}\right)}\cdot \blue{3}^{\log_\blue{3}\left(\purple{\sqrt{3}}\right)}\\&=&\green{6}\purple{\sqrt{3}}\\&=&\blue{3}^{\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)}\end{array}$ and so $\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right) = \log_\blue{3}\left(\green{6} \purple{\sqrt{3}}\right).$

$\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)-\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\frac{\green{b}}{ \purple{c}}\right)\end{array}$

Example

$\begin{array}{lccr}\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)&=&\log_\blue{2}\left(4\right)\end{array}$

Note that $\begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)}&=&\dfrac{\blue{a}^{\log_\blue{a}\left(\green{b}\right)}}{ \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}}\\&=&\dfrac{\green{b}}{\purple{c}}\\&=&\blue{a}^{\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right)}\end{array}$

From this, we conclude that $\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)=\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right).$

Example

$\begin{array}{rcl}\blue{2}^{\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)}&=&\dfrac{\blue{2}^{\log_\blue{2}\left(\green{8}\right)}}{ \blue{2}^{\log_\blue{2}\left(\purple{2}\right)}}\\&=&\dfrac{\green{8}}{\purple{2}}\\&=&\blue{2}^{\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)}\end{array}$ and so $\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)=\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)=\log_\blue{2}\left(4\right).$

$\begin{array}{rcl}\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)&=&\log_{\blue{a}}\left(\green{b}^\purple{n}\right)\end{array}$

Example

$\begin{array}{lccr}\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)&=&\log_\blue{2}\left(\green{4}^\purple{3}\right)\\&=&\log_\blue{2}\left(64\right)\end{array}$

Note that $\begin{array}{rcl}\blue{a}^{\purple{n}\cdot \log_\blue{a}\left(\green{b}\right)}&=&\left(\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\right)^\purple{n}\\&=&\green{b}^\purple{n}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}^\purple{n}\right)}\end{array}$ From this, we conclude that $\purple{n}\cdot \log_\blue{a}\left(\green{b}\right) = \log_\blue{a}\left(\green{b}^\purple{n}\right).$

Example $\begin{array}{rcl}\blue{2}^{\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)}&=&\left(\blue{2}^{\log_\blue{2}\left(\green{4}\right)}\right)^\purple{3}\\&=&\green{4}^\purple{3}\\&=&\blue{2}^{\log_\blue{2}\left(\green{4}^\purple{3}\right)}\end{array}$ and so $\purple{3}\cdot \log_\blue{2}\left(\green{4}\right) = \log_\blue{2}\left(\green{4}^\purple{3}\right)=\log_\blue{2}\left(64\right).$

With the help of these rules we can solve even more questions with logarithms.

Simplify the following expression to a logarithm.

$2-3\cdot\log_{4}\left(2\right)$

$\log_4(2)$

$\begin{array}{rcl} 2-3\cdot\log_{4}\left(2\right)&=&\log_4\left(4^2\right)-3\cdot\log_4\left(2\right)\\ &&\blue{b=\log_a\left(a^b\right)}\\ &=&\log_4\left(16\right)-3\cdot\log_4\left(2\right)\\ &&\blue{\text{calculated $4^2$}}\\ &=&\log_4\left(16\right)-\log_4\left(2^3\right)\\ &&\blue{c\cdot\log_a\left(b\right)=\log_a\left(b^c\right)}\\ &=&\log_4\left(16\right)-\log_4\left(8\right)\\ &&\blue{\text{calculated $2^3$}}\\ &=&\log_4\left(\frac{16}{8}\right)\\ &&\blue{\log_a\left(b\right)-\log_a\left(c\right)=\log_a\left(\frac{b}{c}\right)}\\ \end{array}$

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