How do you calculate #ln(4096)/ln(4)#?
1 Answer
Jun 9, 2016
Explanation:
#ln(4096)/ln(4) = ln(4^6)/ln(4) = (6 color(red)(cancel(color(black)(ln(4)))))/color(red)(cancel(color(black)(ln(4)))) = 6#
Alternatively, use the change of base formula:
#log_a b = (log_c b) / (log_c a)#
to find:
#ln(4096)/ln(4) = log_4 4096 = log_4 4^6 = 6#
