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Mathematics 16 Online
OpenStudy (anonymous):

Use induction to prove the following: 3^n > 2^n for all positive integers

OpenStudy (anonymous):

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

ganeshie8 (ganeshie8):

3^k > 2^k multiply 3 both sides 3^(k+1) > 3*2^k > 2*2^k > 2^(k+1) QED

OpenStudy (anonymous):

Where does the three go on the right side?

ganeshie8 (ganeshie8):

\[\large a \gt 3 \implies a \gt 2\]

ganeshie8 (ganeshie8):

\[\large a \gt 3*2^k \implies a \gt 2*2^k\]

OpenStudy (anonymous):

Okay that make more sense. Thanks!

ganeshie8 (ganeshie8):

yw!

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