How do you simplify #64^(log_4 (8y))#?

1 Answer
Dec 31, 2015

Answer:

#512y^3#

Explanation:

#64=4^3#, so the expression can be written as

#=>(4^3)^(log_4(8y))#

#=>4^(3log_4(8y))#

Rewrite using the rule: #alog_b(c)=log_b(c^a)#

#=>4^(log_4((8y)^3))#

The #4# and the #log_4# will cancel since exponentiation and logarithmic functions are inverses.

  • #a^(log_a(b))=b#

#=>(8y)^3#

#=>512y^3#