# How do you evalute Log_ 3 (18)?

Jan 1, 2016

${\log}_{3} \left(18\right) = 2 + {\log}_{3} \left(2\right) \approx 2.6309$

#### Explanation:

${\log}_{3} \left(18\right) = {\log}_{3} \left({3}^{2} \times 2\right)$

Use the rule: ${\log}_{a} \left(b c\right) = {\log}_{a} \left(b\right) + {\log}_{a} \left(c\right)$

$\implies {\log}_{3} \left({3}^{2}\right) + {\log}_{3} \left(2\right)$

Use the rule: ${\log}_{a} \left({a}^{b}\right) = b$

$\implies 2 + {\log}_{3} \left(2\right)$

This is a simplified answer. However, if you want your answer in decimal form, use a calculator. Since many calculators don't have the capability of finding logarithms with a specific base, use the change of base formula.

${\log}_{a} \left(b\right) = \frac{{\log}_{c} \left(b\right)}{{\log}_{c} \left(a\right)}$

Thus, since the $\ln$ button is on most calculators (or just the $\log$ button),

${\log}_{3} \left(2\right) = \ln \frac{2}{\ln} \left(3\right) = \log \frac{2}{\log} \left(3\right) \approx 0.6309$

Thus,

${\log}_{3} \left(18\right) = 2 + {\log}_{3} \left(2\right) \approx 2.6309$