How do you solve #4log_4 7#?

1 Answer
May 7, 2015

You can take the #4# in front of the #log# "inside" it as:
#log_4(7)^4=#
#log_(4)(2401)=#
You can change base to use natural logarithms, that can be evaluated using a pocket calculator, as:
#log_(4)(2401)=ln(2401)/ln(4)=5.614#