very very easy

yeah cele is rite...some cond is missing

WARNING:Long method.....(but did it just to check validity of question)

considering m,n>0
take log both sides
mn.log2>nlogm
=> n(mlog2-logm)>0
Since n>0 so (mlog2-logm) must be greater than 0

consider f(m)=p=mlog2-logm
p'=log2-1/m
=>p changes sign at m=1/log2

so the question should be true for m>1/log2 and n>0

See Easy
2^mn>mⁿ
rite
Now you cancel n on both the sides
Therefore
2^m>m
This is understandable therefore done

waaaaaaaahhhhh virang.u posted while i was typing.....

2^m>m
mlog2>logm
m>logm/log2

or by taking

2>m^1/m
then take AM GM inequalities..

it boils down to

2^m>m

now m=1 is true

consider a function 2^m -m f'(x) = ln2.2^m - 1

this f'(x) > 0 for mε N

so its a strictly inc function

has m=1 is already true
its true for all mε N

or.................
we cud do by PMI
2^m>m
for f(1) 2>1 so f(k) is true
let f(k) be true so..
2^k>k
now we need to prove f(k+1) is true
LHS..2^k+1=2^k.2
>k.2
>K+1

so f(k+1) is also true

so by induction...

2^m>m

easiest proof:

consider a set A having n distinct elements and set B having m distinct elements

No. of relations from A→B = 2^mn
No. of functions from A→B = mⁿ

As No. of relations < No. of functions so 2^mn>mⁿ
similarly taking A has m elements and B has n elements it can be proved that
2^mn>n^m

eureka sorry i dint see that

#3 is = #10