IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

You must note that the proofs you suppy should be conclusive. There should be nothing left hanging in the air to be guessed or surmised.

Comparing $99^{100}$ and $100^{99}$ is equivalent to comparing $\sqrt [99] {99}$ and $\sqrt [100] {100}$

Now consider the function $f(x) = \sqrt [x] {x}$

$f'(x) = \frac{1 - \log x}{x^2}$

Hence for x>1, f is a decreasing function.

Thus, we deduce that $\sqrt [99] {99} > \sqrt [100] {100}$ .