How do you calculate #log _6 (1)#?

1 Answer
Jun 20, 2016

#log_6(1)=color(green)(0)#

Explanation:

Remember what #log# actually means:
#color(white)("XXX")log_b(a)=color(red)(c)# means #b^color(red)(c)=a#

In the given case
#color(white)("XXX")#if #log_6(1)=color(red)(c)# this means that #6^color(red)(c)=1#

...and since #6^color(red)(0) =1#

#log_6(1)=color(red)(0)#