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4 ^ (k + 1) – 4 ^ k

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4 ^ (k + 1) – 4 ^ k

by fltingley » Mon Apr 20, 2009 5:46 am
Can someone walk me through how 4 ^ (k + 1) – 4 ^ k factors out to 4^k(4-1)?

I see it when I substitute numbers (e.g..- k =2), but the algebraic concept is baffling me.

Guess I just need someone to translate this into English.

Thanks!
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by mals24 » Mon Apr 20, 2009 6:10 am
Remember these 2 exponential rules:

1. A^x*A^y = A^(x+y)

This means when you multiply identical bases (A), you add their exponents (x+y).

2. (A^x)/(A^y) = A^(x-y)

This means when you divide identical bases (A), you subtract their exponents (x-y).

4^(k+1)

Applying rule 1: 4^k*4

4^(k+1) - 4

4^k*4 - 4

Take out 4 since it is common

4(4^k-)...which is what your solution is.
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