# Please help me solve this question: Prove that 2^n > 3n for all positive integers n `>=` 4 Thanks...

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### 1 Answer

`2^n>3n`

This can be solved by mathematical induction.

It has given that n>=4.

So the lowest value of n=4

`n=4`

`2^4 = 16`

`3xx4 = 12`

`16>12`

So when n = 4 the result is true.

Let us assume when n = p where p>4 the result is true.

`2^p>3p`

Now we have to prove that when `n = p+1` the result is true.

`2^p>3p`

`2^pxx2>3pxx2`

`2^(p+1)>6p`

What we have to show that is `2^(p+1)>3(p+1)` .

But what we have obtained is `2^(p+1)>6p` .

So if we can prove `6p>3(p+1)` then we have say;

`2^(p+1)>6p>3(p+1)`

So ultimately we will endup with `2^(p+1)>3(p+1)`

Let us assume `6p>3(p+1)`

`6p>3(p+1)`

`6p-3(p+1) > 0`

`3p-3>0`

`3(p-1) > 0`

We know that `p>4` . Then `(p-1)>0` and `3(p-1) > 0` .

This means our assumption is correct.

`6p>3(p+1)`

Now we can say `2^(p+1)>3(p+1)`

So for n = p+1 the result is correct.

*Therefore from mathematical induction for all `n>=4` ;*

`2^n>3n`

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