Let
$\text{\hspace{0.17em}}m={\mathrm{log}}_{b}M\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}n={\mathrm{log}}_{b}N.\text{\hspace{0.17em}}$ In exponential form, these equations are
$\text{\hspace{0.17em}}{b}^{m}=M\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{b}^{n}=N.\text{\hspace{0.17em}}$ It follows that
Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider
$\text{\hspace{0.17em}}{\mathrm{log}}_{b}(wxyz).\text{\hspace{0.17em}}$ Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
For quotients, we have a similar rule for logarithms. Recall that we use the
quotient rule of exponents to combine the quotient of exponents by subtracting:
$\text{\hspace{0.17em}}{x}^{\frac{a}{b}}={x}^{a-b}.\text{\hspace{0.17em}}$ The
quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.
Given any real number
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and positive real numbers
$\text{\hspace{0.17em}}M,$$N,$ and
$\text{\hspace{0.17em}}b,$ where
$\text{\hspace{0.17em}}b\ne 1,$ we will show
Let
$\text{\hspace{0.17em}}m={\mathrm{log}}_{b}M\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}n={\mathrm{log}}_{b}N.\text{\hspace{0.17em}}$ In exponential form, these equations are
$\text{\hspace{0.17em}}{b}^{m}=M\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{b}^{n}=N.\text{\hspace{0.17em}}$ It follows that
For example, to expand
$\text{\hspace{0.17em}}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right),$ we must first express the quotient in lowest terms. Factoring and canceling we get,