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!
4 ^ (k + 1) – 4 ^ k
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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.
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.