For valid bases a,c and b>0, log_a b= log_c blog_c a. The order is argument over base. Why: put u=log_a b, so au=b. Taking logarithms to base c gives ulog_c a=log_c b. Since a 1, log_c a 0, and division gives the formula. Method: choose a common convenient base, usually e, 10, or a supplied symbolic base; replace each logarithm by a quotient; simplify before calculating. For example, log_8 4= log_2 4log_2 8= 23. Also 1log_7 e=ln7. Exam check: log_a b and log_b a are reciprocals only when both logarithms are defined; the reciprocal form requires b 1. Do not invert the change-of-base quotient.

For valid bases \(a,c\) and \(b>0\), \(\log_a b=\dfrac{\log_c b}{\log_c a}\). The order is argument over base.

Why: put \(u=\log_a b\), so \(a^u=b\). Taking logarithms to base \(c\) gives \(u\log_c a=\log_c b\). Since \(a\ne1\), \(\log_c a\ne0\), and division gives the formula.

Method: choose a common convenient base, usually \(\mathrm e\), \(10\), or a supplied symbolic base; replace each logarithm by a quotient; simplify before calculating.

For example, \(\log_8 4=\dfrac{\log_2 4}{\log_2 8}=\dfrac23\). Also \(\dfrac1{\log_7\mathrm e}=\ln7\).

Exam check: \(\log_a b\) and \(\log_b a\) are reciprocals only when both logarithms are defined; the reciprocal form requires \(b\ne1\). Do not invert the change-of-base quotient.

Why change of base works

Why change of base works