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 $\log_{10}{8} + \log_{10}{5} + \log_{10}{25}$

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

## 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:

$\begin{aligned} \log_2{x} + 6\,\log_8{y} - \log_4{z} &= \log_2{x} + 6\frac{\log_2{y}}{\log_2{8}} - \frac{\log_2{z}}{\log_2{4}} \\ &= \log_2{x} + \frac{6\,\log_2{y}}{3} - \frac{\log_2{z}}{2} \\ &= \log_2{x} + 2\,\log_2{y} - \frac{\log_2{z}}{2} \\ &= \log_2{x} + \log_2{y^2} - \log_2{\sqrt{z}} \\ &= \log_2{\left( \frac{xy^2}{\sqrt{z}} \right )} _\square \end{aligned}$

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