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

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Answer 1 · 2015-12-31T06:50:15

#512y^3#

#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#