Prove x^n < n^n 2^x

Laila Murphy

Laila Murphy

Answered question

2022-11-04

Prove x n < n n 2 x
Given that
x < 2 x
is always true, use it to prove that
x n < n n 2 x
Here are the steps that I've taken so far:
Reduce
x < 2 x
to
log ( x ) < x
Then
x n < n n 2 x
n log ( x ) < n log ( n ) + x
log ( x ) < log ( n ) + x n
And that's where I got stuck.
If it were
log ( x ) < log ( n ) + x
the proof would be self evident, because since
log ( x ) < x
it would be obvious that ( x + anything else ) would be more than log ( x ), but in this case it isn't.
Is there something else that I'm missing? By direct calculation my last step would be true, but how do I relate it to the first statement?
Any help would be greatly appreciated, guys. Thanks a lot!

Answer & Explanation

x713x9o7r

x713x9o7r

Beginner2022-11-05Added 15 answers

Let x 0. The relation x n < n n 2 x can be rewritten as ( x n ) n < 2 x and then as x n < 2 x / n . But we are told that t < 2 t for all t. Put t = x n
Remark: Let x = 100 and n = 2. Then it is not true that x n < n n 2 x

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