TITLE: compare $m=50^{50}$ with $n=49^{51}$
QUESTION [8 upvotes]: A multiple choice question:

If $m=50^{50}$ and $n=49^{51}$, then
(A) $m>n$
(B) $m<n$
(C) $m=n$
(D) The given information is not enough

My attempt:
Since ordinary calculators can not evaluate large numbers as $m$ and $n$, then we can use a trick, which is taking the logarithm of both $m$ and $n$ to the same base, lets use $\ln$ (log to the base $e$).
$50\ln(50)$ VS $51\ln(49)$
$195.60$ VS $198.48$
Hence $49^{51}$ is greater.
So, B must be the correct choice.

This question was asked in a national exam for high school students.

However:

Calculators are not allowed.
Log tables are not provided.
Students may not have any knowledge about logarithms and their properties.
Students should have basic knowledge about exponents like $(a/b)^k=a^k/b^k$, $a^j \times a^k = a^{(j+k)}$, and some other basics.
The average time to solve a question in this exam is 75 seconds.


How can we answer this question?

REPLY [4 votes]: In this answer, it is shown that $\left(1+\frac1{n-1}\right)^n$ is decreasing. That means that its reciprocal $\left(1-\frac1n\right)^n$ is increasing. Thus, for $n\ge2$, we have
$$
\left(1-\frac1n\right)^n\ge\frac14\tag1
$$
Therefore,
$$
\begin{align}
\frac{49^{51}}{50^{50}}
&=49\left(1-\frac1{50}\right)^{50}\\
&\ge\frac{49}4\tag2
\end{align}
$$

Another Proof of $\boldsymbol{(1)}$
Using Theorem $1$ from this answer with $m=2$, we get
$$
\begin{align}
\left(1-\frac1n\right)^n
&\ge1-\frac{n}{n}+\frac{n(n-1)}{2n^2}-\frac{n(n-1)(n-2)}{6n^3}\\
&=\frac{n^2-1}{3n^2}\tag3
\end{align}
$$
For $n\ge2$, $(3)$ gives $(1)$.