# Evaluating Logarithmics

• Dec 11th 2007, 06:40 PM
Macleef
Evaluating Logarithmics
$log_26 + log_{28} - log_23$

How do I put them into common bases? I was thinking of using power law, but there are no exponents

• Dec 11th 2007, 06:46 PM
kalagota
Quote:

Originally Posted by Macleef
$log_26 + log_{28} - log_23$

How do I put them into common bases? I was thinking of using power law, but there are no exponents

i assume, the question should be $\log_2 6 + \log_2 8 - \log_2 3$

if so, note that $\log_a b^x = x \log_ab$ and $\log a + \log b = \log ab$, $\log a - \log b = \log \frac{a}{b}$

thus we have $\log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3$
$= 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4$..
• Dec 11th 2007, 06:47 PM
Macleef
Quote:

Originally Posted by kalagota
i assume, the question should be $\log_2 6 + \log_2 8 - \log_2 3$

if so, note that $\log_a b^x = x \log_ab$ and $\log a + \log b = \log ab$, $\log a - \log b = \log \frac{a}{b}$

thus we have $\log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3$
$= 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4$..

I think you're right, there must be a typo within the question in the textbook