Evaluating Logarithmics

**Macleef** · December 11th 2007, 06:40 PM

$log_26 + log_{28} - log_23$

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

Answer: 4

**kalagota** · December 11th 2007, 06:46 PM

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

Answer: 4

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

**Macleef** · December 11th 2007, 06:47 PM

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

Thread: Evaluating Logarithmics

LinkBack

Thread Tools

Display