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Use induction to prove the following: 3^n > 2^n for all positive integers
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I have a base case of 1 making 3^1>2^1 true and have n=k, making 3^k>2^k also true. n=k+1 seems like it would be a simple plug in, but I feel like I'm missing something
3^k > 2^k multiply 3 both sides 3^(k+1) > 3*2^k > 2*2^k > 2^(k+1) QED
Where does the three go on the right side?
\[\large a \gt 3 \implies a \gt 2\]
\[\large a \gt 3*2^k \implies a \gt 2*2^k\]
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Okay that make more sense. Thanks!
yw!
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