The following two equations are equivalent:
\[ b^y = x \qquad \text{and} \qquad y = \log_b{x}. \] Here, \(y\) is the **logarithm** of \(x\) to the base \(b\). In other words, \(b\) needs to be raised to the power of \(y\) to equal \(x\).

Logarithms are only defined when \(b\) is positive and not equal to \(1\), and \(x\) is positive.

Note that \[ b^{\log_b{x}} = x \text{ and } \log_b{b^y} = y.\] Two of the most common log bases you will encounter are \(\log_{10}\) and \(\log_{e}\) (\(e\) is Euler's number). They are sometimes written as simply \(\lg\) and \(\ln\), respectively.

Several important identities are used to relate logarithms to one another. They are known as **logarithmic laws**.

\[ \begin{array}{lrcl} \textbf{Products: } & \log_b{(xy)} &=& \log_b{(x)} + \log_b{(y)} \\ \\ \textbf{Ratios: } & \log_b{ \left(\frac{x}{y}\right) } &=& \log_b{(x)} - \log_b{(y)} \\ \\ \textbf{Powers: } & \log_b{(x^p)} &=& p\,\log_b{(x)} \\ \\ \textbf{Roots: } & \log_b{\left(\sqrt[p]x \right )} &=& \frac{\log_b{(x)}}{p} \\ \\ \textbf{Change of bases: } & \log_b{(x)} &=& \frac{\log_a{(x)}}{\log_a{(b)}} \end{array} \]

Here are a couple of example problems:

## Evaluate \( \log_{10}{8} + \log_{10}{5} + \log_{10}{25} \)

\[ \begin{align} \log_{10}{8} + \log_{10}{5} + \log_{10}{25} &= \log_{10}{\left( 8 \cdot 5 \cdot 25 \right)} \\ &= \log_{10}{1000} \\ &= 3 \, _\square \end{align} \]

## Solve for \(x\): \( \log{x} = 1 - \log{\left(x+3\right)} \)

Use the laws of logarithms to put \(x\) into a single logarithm.

\[ \log{x} + \log{(x+3)} = 1 \\

\log{\left ( x(x+3) \right )} = 1\] This logarithmic expression can then be rewritten as an exponential.

\[ x(x+3) = 10^1 \] This gives us the quadratic equation \( x^2 + 3x - 10 = 0 \), which has the roots \( x = 2 \) and \( x = -5 \).We immediately have to reject \( x = -5 \), because \( \log{(-5)} \) and \( \log{(-5+3)} \) are negative and, therefore, undefined. Thus, the correct answer is \( x = 2. \, _\square \)

Here is an application that uses all five logarithmic laws:

## Rewrite \( \log_2{x} + 6\,\log_8{y} - \log_4{z} \) as a single logarithm:

\[ \begin{align} \log_2{x} + 6\,\log_8{y} - \log_4{z} &= \log_2{x} + 6\frac{\log_2{y}}{\log_2{8}} - \frac{\log_2{z}}{\log_2{4}} \\

&= \log_2{x} + \frac{6\,\log_2{y}}{3} - \frac{\log_2{z}}{2} \\

&= \log_2{x} + 2\,\log_2{y} - \frac{\log_2{z}}{2} \\

&= \log_2{x} + \log_2{y^2} - \log_2{\sqrt{z}} \\

&= \log_2{\left( \frac{xy^2}{\sqrt{z}} \right )} _\square \end{align} \]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

There are no comments in this discussion.