Precalculus Review

Laws of Logarithms

Let $v = \log_a b$ and $w = \log_a c$ . Then we can verify the following lows of logarithms:

$\log_a(bc) = \log_a b + \log_a c$ ( since $a^y = a^{v + w} = a^v \cdot a^w = bc$ );
$\log_a\left(\frac{ b }{c}\right) = \log_a b - \log_a c$ ( since $a^y = a^{v - w} = a^v \cdot a^{-w} = \dfrac{ b }{c}$ );
$\log_a b^k = k \log_a b$ ( since $a^y = a^{kv} = \left(a^v\right)^k = b^k$ ).

Change of base. Observe that $u = \log_b c$ implies . Taking $\log_a$ in the both side of the equation , we have $\log_a c = \log_a b^u = u \log_a b$ . This implies that

$\displaystyle u = \log_b c = \frac{\log_a c}{\log_a b} .$