IIT JEE , AIEEE Forum - Mathematics , Algebra

rudra.panda

Which one is larger?100⁹⁹or 99^100.

Please give the solution.

Vincent

The second one

allamraju

sarvpriye

(99)¹⁰⁰> (100)⁹⁹

Because,

(100)⁹⁹= (99+1)⁹⁹

= ⁹⁹C₀ (99)⁹⁹+ ⁹⁹C₁(99)⁹⁸+ ....................... + 1

= (99)⁹⁹ + (99).(99)⁹⁸ +............... + 1

= 2.(99)⁹⁹+ terms relatively much smaller terms

and this is much smaller than 99.(99)⁹⁹i.e (99)¹⁰⁰

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 $\sqrt [99] {99} > \sqrt [100] {100}$ .

animal

thanx hsbhatt sir

plz give us some more tips like this.

karansingh

excuse me hsbhatt sir,
it is decresing function for x > 10.
and not x > 1.

plz say where am i wrong.

hsbhatt

Thanx for pointing out the error. It is a decreasing function for x>e and not x>1 as i posted before

Here the log is taken to the base e

So, if x>e, then $\log_e {x} > \log_e{e} = 1$

Hence $1- \log_e{x} < 0$ for x>e