Why does #lna – lnb = ln(a/b)#?
Why does #lna – lnb = ln(a/b)#? It does not matter what base we use providing the same base is used for all logarithms, here we are using bease #e#. Cerchiamo di definire #A,B.C# as follows=: # A = ln a iff a = e^A #, # B = ln b iff b = e^B … Leggi tutto