Solution 3.3:3b
From Förberedande kurs i matematik 1
(Difference between revisions)
m (Lösning 3.3:3b moved to Solution 3.3:3b: Robot: moved page) |
|||
Line 1: | Line 1: | ||
- | {{ | + | Because we are working with |
- | < | + | <math>\log _{9}</math>, we express |
- | {{ | + | <math>{1}/{3}\;</math> |
+ | as a power of | ||
+ | <math>\text{9}</math>, | ||
+ | |||
+ | |||
+ | <math>\frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}</math> | ||
+ | |||
+ | |||
+ | Using the logarithm laws, we get | ||
+ | |||
+ | |||
+ | <math>\log _{9}\frac{1}{3}=\log _{9}9^{-{1}/{2}\;}=-\frac{1}{2}\centerdot \log _{9}9=-\frac{1}{2}\centerdot 1=-\frac{1}{2}.</math> |
Revision as of 14:06, 25 September 2008
Because we are working with \displaystyle \log _{9}, we express \displaystyle {1}/{3}\; as a power of \displaystyle \text{9},
\displaystyle \frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}
Using the logarithm laws, we get
\displaystyle \log _{9}\frac{1}{3}=\log _{9}9^{-{1}/{2}\;}=-\frac{1}{2}\centerdot \log _{9}9=-\frac{1}{2}\centerdot 1=-\frac{1}{2}.