Software Developer

$log_aM +  log_aN = log_aMN$  (1) proof:  assume: $log_aM = m, log_aN = n$  so:  $a^m = M, a^n = N \Rightarrow M \cdot N = a^m \cdot a^n = a^{m+n}$  $log_a{MN} = m + n = log_aM + log_aN$  $log_aM - log_aN = log_a\frac{M}{N}$ (2) proof: assume: $log_aM = m, log_aN = n$ so: $a^m = M, a^n = N \Rightarrow \frac{M}{N}=\frac{a^m}{a^n} = a^{m - n} \Rightarrow log_a{\frac{M}{N}} = m - n = log_aM - log_aN$ $log_aa^M = M$ (3) proof: assume: $a^M = B \Rightarrow log_aB = M$ so: $log_aa^M = M$ $a^{log_aM} = M$ (4) proof: assume: $log_aM = B$ so: $a^B = M$ $\because B = log_aM, a^B = M$ $\therefore a^B = a^{log_aM} = M$ $log_aM^N = Nlog_aM$ (5) proof: $log_aM^N = log_a(\overbrace{M \cdot M \cdots M}^{N})$ $\because log_aM +  log_aN = log_aMN$  property (1) $\therefore log_a(\overbrace{M \cdot M \cdots M}^{N}) = \overbrace{log_aM + log_aM + \cdots log_aM}^{N} = Nlog_aM$ $log_ab = \frac{log_cb}{log_ca}= \frac{lnb}{lna} = \frac{lgb}{lga}$ (6) ...