If #1/64 = 4^(2s-1)*16^(2s+2)#, what is the value of #s#?

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Answer 1 · 2017-04-26T02:47:38

#s = -1#.

Write in common bases.

#2^(-6) = (2^2)^(2s - 1) * (2^4)^(2s + 2)#

Use #(a^n)^m = a^(nm)#:

#2^(-6) = 2^(4s - 2) * 2^(8s + 8)#

Use #a^n * a^m = a^(n + m)#.

#2^(-6) = 2^(4s - 2 + 8s + 8)#

#2^(-6) = 2^(12s + 6)#

You can eliminate bases now.

#-6 = 12s + 6#

#-12 = 12s#

#s = -1#

Hopefully this helps!