$\displaystyle log_26 + log_{28} - log_23$
How do I put them into common bases? I was thinking of using power law, but there are no exponents
Answer: 4
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 $\displaystyle \log_2 6 + \log_2 8 - \log_2 3 = \log_2 6 + \log_2 2^3 - \log_2 3$
$\displaystyle = 3\log_2 2 + \log_2 6 - \log_2 3 = 3 + \log_2 \frac{6}{3} = 3 \log_2 2 = 3 + 1 = 4$..