Posted by jack on Tuesday, March 11, 2008 at 10:45pm.
4^(n+1) = 2^[2*(n+1)]= 2^(2n+2)
8^(n-2)= (2^3)^(n-2) = 2^(3n-6)
If the 8^(n-2) is supposed to be in a denominator,
2^n * 4^(n+1)
____________
8^(n-2)
= 2^(n + 2n + 2 -3n +6)
= 2^8 = 256
no matter what n is. Pick a value of n and prove it for yourself.
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