# agebra

Hey, im a little confused with this question:

2^n x 4^(n+1)
/
8^(n-2)

I have to change all bases to 2 then simplify fully.

Thanks everoyne.

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1. 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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