How do you solve #16^ { - 2v } = 64^ { 2v }#?

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Answer 1 · 2018-07-28T09:52:42

#v=0#

#16^(-2v)=64^(2v)#

#(4^2)^(-2v)=(4^3)^2v#

#4^(-4v)=4^(6v)#

#-4v=6v#

#-4v-6v=0#

#-10v=0#

#v=0#

Answer 2 · 2018-07-28T10:13:38

#v=0#

#"express 16 and 64 in base 4"#

#(4^2)^(-2v)=(4^3)^(2v)#

#4^(-4v)=4^(6v)#

#"equating the exponents gives"#

#-4v=6vrArr2v=0rArrv=0#