Laws of Logarithms

For any positive base b ≠ 1 and any positive numbers x, y, and n:

$\log_{b} (x y) = \log_{b} x + \log_{b} y_{}$
$\log_{b} (\frac{x}{y}) = \log_{b} x - \log_{b} y$
$\log_{b} x^{n} = n \log_{b} x$
$\log_{b} b^{n} = n$ (In particular, $\log_{b} 1 = \log_{b} b^{0} = 0$ and $\log_{b} b = \log_{b} b^{1} = 1$ .)

For example, if log_b 2 = x and log_b 3 = y, then:

\begin{array}{l} \log_{b} 6 & = & \log_{b} (2 \times 3) = \log_{b} 2 + \log_{b} 3 = x + y \\ \log_{b} \frac{2}{3} & = & \log_{b} 2 - \log_{b} 3 = x - y \\ \log_{b} 72 & = & \log_{b} (8 \times 9) = \log_{b} 8 + \log_{b} 9 \\ = & \log_{b} 2^{3} + \log_{b} 3^{2} \\ = & 3 \log_{b} 2 + 2 \log_{b} 3 \\ = & 3 x + 2 y \end{array}

As you will see in Section 2-E, the third rule in KEY FACT A18 allows you to solve equations in which the variable is an exponent by bringing the variable down to the base line. To solve the equation 2^x = 512, for example, take the logarithm of both sides: log_b 2^x = log_b 512. Then

\log_{b} 512 = \log_{b} 2^{x} = x \log_{b} 2 \Rightarrow x = \frac{\log_{b} 512}{\log_{b} 2}

Note that what you choose for the base b does not matter. As a consequence, you might as well let b = 10. Therefore, $x = \frac{\log 512}{\log 2}$ . Now use your calculator:

x = (\log 512) \div (\log 2) = 9