How do you calculate #log_4 (1/16) #?

Question

5994 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-05T07:05:29

#log_4(1/16)=?#

#log_a b=x#

#b=a^x#

#log_4(1/16)=k#

#1/16=4^k#

#1/(4^2)=4^k#

#color(green)(4)^(-2)=color(green)(4)^k#

#k=-2#

#log_4(1/16)=-2#