let's say we had an equation that looked like 10 to the power of x equals 7. how would you solve that well traditionally without using log you have no way of getting X by itself but if you remember ...

\({\log _a}a = 1\) (since \({a^1} = a\)) so \({\log _7}7 = 1\) \({\log _a}1 = 0\) (since \({a^0} = 1\)) so \({\log _{20}}1 = 0\) \({\log _a}p + {\log _a}q = {\log _a ...