$\displaystyle log_26 + log_{28} - log_23$

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

Answer: 4

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- Dec 11th 2007, 06:40 PMMacleefEvaluating Logarithmics
$\displaystyle log_26 + log_{28} - log_23$

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

Answer: 4 - Dec 11th 2007, 06:46 PMkalagota
i assume, the question should be $\displaystyle \log_2 6 + \log_2 8 - \log_2 3$

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

thus we have $\displaystyle \log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3$

$\displaystyle = 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4$.. - Dec 11th 2007, 06:47 PMMacleef