What is the greater number? [duplicate]

Question

I would like to prove that $100^{99}<99^{100}$.

My attempt:

$\dfrac{100^{99}}{99^{99}}<99$

$\left(\dfrac{100}{99}\right)^{99}<99$

$\left(\dfrac{2^2\cdot5^2}{3^2\cdot11}\right)^{99}<99$

$\dfrac{2^{198}\cdot5^{198}}{3^{198}\cdot11^{99}}<99$

But now I am stuck. Please can you help me?

@lulu Good year. Do leave my answer here or put it in the link? — Sebastiano, Commented Jan 1, 2023 at 19:57
@Sebastiano Happy New Year to you as well. I see no problem leaving the answer where it is, though my guess is that logs weren't meant to be involved in the solution here. At least not numerical logs. Usually this sort of problem is solved by looking at general properties of the function $F(x)=\frac {\ln x}x$. — lulu, Commented Jan 1, 2023 at 20:00
When $x$ and $y$ are both larger than $2.718{\dots}$, it turns out that $x^y>y^x$ iff $y>x$. EDIT: Here's a graph. The region where $x^y>y^x$ is shaded. i.imgur.com/sI65P9Y.jpg — Akiva Weinberger, Commented Jan 1, 2023 at 20:04
@PeterPhipps It's an exact duplicate, with multiple highly upvoted answers. In my view, this question should be closed either as a duplicate or because it, too, shows a lack of serious effort. In my view, closing it as a duplicate is more informative (at least the OP gets their answer)...not sure what the official view on that is. — lulu, Commented Jan 1, 2023 at 20:13

Sebastiano · Accepted Answer · 2023-01-01 20:09:42Z

2

$\log$ is a strictly increasing function on positive reals.

Let $x = 99^{100}$ and $y=100^{99}$; taking logarithm to the base $10$,

$$\log x = 100 \log 99 = 100\cdot 1.9956= 199.56$$

$$\log y= 99 \log 100= 99\cdot 2 = 198$$

Since $\log x > \log y$, $x$ is greater than $y$.

answered Jan 1, 2023 at 19:56

Sebastiano

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