View Full Version : Logarithms
amyyzhu
Jul 9, 2012, 11:18 PM
How do you simplify logs with fractions or roots? With normal simple numbers, I understand but using fractions or roots, I get really confused. I think it's also because I'm rocky with indices as a whole to start with so that didn't help.
Example;
Given that logb(2) = 0.3010 and logb(3) = 0.4771, find: logb(root 2) & logb(1/3)
Thank you in advance!
ebaines
Jul 10, 2012, 06:29 AM
As you study logarithms you really must memorize and become comfortable with the key relationships, which include:
\log _a (bc) = \log_a b + \log_a c\\
\log _a (\frac b c } = \log _a b - \log _a c\\
\log _a b^n = n \log _a b\\
\log_a b \ \times\ \log _b c = \log _a c\\
a^{\log_a b} = b
Using the third relationship on your first example you have:
\log_b \sqrt 2 = \log _b 2^{1/2} = \frac 1 2 \log_b 2
For the log of 1/3 you can do it two different ways. First using the relationship for \log(a^n) again:
\log_b \frac 1 3 = \log_b 3 ^{-1} = -1 \log _b 3
Or using the relationship for \log (a/b):
\\log_b \frac 1 3 = \log_b 1 - \log_b 3.
Since \log 1 = 0, this leads to the same result.