AOD

Q100 ans 1


@sanketbiradar
Why for a>1 a^x will never intersect y=x
Please prove this by calculus

@nishantthakre
consider ln xx . get its monotonic behavr.
now e^x is mirror image in inverse f of ln x

This post is deleted!

@ashishdeosingh
But here it is a and not e
I didn't understand
Suppose we have to prove for monotonicity of it's inverse as derivative of both have same sign
To prove monotonicity of log_a_xx
=(logx/loga) x
f'(x)=1/xloga 1
f"(x)=1/x^2loga
Always ve
f'(x)>0
For positive x
xloga<1
It depends on x and a both
How to do it

a>1
e>1
Same behaviour

@kirito @ashishdeosingh
ok we can predict it by analyzing graph of e^x
x for a>e it will be of course true
But what about a=2 which is less than e

@nishantthakre
any thing greater than 1 will hv same behavoiur

@ashishdeosingh
No
I am uploading a computer created graph
For a^x and x
For some values e.g. a=1.3 graphs intersect
After a certain value of a they do not
We should derive mathematically for what values of a it is above y=x

@NishantThakre only integral values asked

@sanketbiradar
Then we must show for a=2 how it is above y=x always( separately from e
I hope I am able to convey what I meant)
If we prove e^x is always above y=x
Then it is obviously true for a>e as a^x>e^x
But 2 is the only case when 2<e
We must show that 2^x lies above y=x separately

