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

$v=0$

$"express 16 and 64 in base 4"$

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

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

$"equating the exponents gives"$

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

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