Change of base formula

Revision as of 19:07, 7 September 2019 by Awesomedude86 (talk | contribs) (Proof 2)

The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases.

For any positive real numbers $d,a,b$ such that neither $d$ nor $b$ are $1$, we have

\[\log_b a = \frac{\log_d a}{\log_d b}.\]

This allows us to rewrite a logarithm in base $b$ in terms of logarithms in any base $d$. This formula can also be written

\[\log_b a \cdot \log_d b = \log_d a.\]

Proof

Let $\log_b a = y$.

Then $b^y = a$.

And, taking the $\log_d$ of both sides, we get \[\log_d (b^y) = \log_d (a)\]

By the properties of logarithms, \[y\log_d b = \log_d a\]

Substituting for y, \[\log_b a \cdot \log_d b = \log_d a\]

Use for computations

The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that

\[\log_{\frac{1}{4}} 32\sqrt{2} = \frac{\log_2 32\sqrt{2}}{\log_2 \frac{1}{4}} = \frac{\frac{11}{2}}{-2} = -\frac{11}{4}.\]

The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate $\log_{7} 19$, you would first convert it to the form $\frac{\log_{10}{19}}{\log_{10}{7}}$. Then you would evaluate it using the base-10 log function on the calculator.

Special cases and consequences

Many other logarithm rules can be written in terms of the change of base formula. For example, we have that $\log_b a = \frac{\log_a a}{\log_a b} = \frac{1}{\log_a b}$. Using the second form of the change of base formula gives $\log_b a^n = \log_b a \cdot \log_a a^n = n \log_b a$.

One consequence of the change of base formula is that for positive constants $a, b$, the functions $f(x) = \log_a x$ and $g(x) = \log_b x$ differ by a constant factor, $f(x) = (\log_a b) g(x)$ for all $x > 0$.


This article is a stub. Help us out by expanding it.