compare $m=50^{50}$ with $n=49^{51}$ 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?
 A: Use Bernoulli's inequality with $x=-\frac{1}{50}$ and $r=48$:
$$\frac{49^{51}}{50^{50}}=\frac{49^3}{50^2}\cdot\frac{49^{48}}{50^{48}}=\frac{49^3}{50^2}\left(1-\frac{1}{50}\right)^{48}\ge\frac{49^3}{50^2}\left(1-\frac{48}{50}\right)=\frac{2\cdot49^3}{50^3}>1$$
A: 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 $(1)$.
A: Hint: Use the inequality $$\left(1-\frac{1}{x}\right)^x>\frac{1}{x-1}$$
A: We have $$\frac{n}{m} = \frac{49^{51}}{50^{50}} = 49 \cdot \left(\frac{49}{50}\right)^{50} = 49 \cdot \left(1-\frac{1}{50}\right)^{50} \approx \frac{49}{e} > 1$$
Even if you dont know that for large $n$ $$ \left(1-\frac{1}{n}\right)^{n} \approx \frac{1}{e}$$
as long as you're able to tell that $$\left(1-\frac{1}{50}\right)^{50} > \frac{1}{49}$$ you're fine. You can get that for example from Bernoulli's inequality. For $x>-1$:
$$ (1+x)^n \ge 1+xn$$
so
$$ \left(1-\frac{1}{50}\right)^{50} = \left(\left(1-\frac{1}{50}\right)^{25} \right)^2 \ge \left(1-\frac{1}{50}\cdot 25\right)^2 = \left(\frac12\right)^2 = \frac14 > \frac{1}{49}$$
A: Knowing the following fairly common limit is very useful:
$$ \lim_{x \rightarrow \infty}\left(1+\frac{1}{x} \right)^x=e \approx 3$$
$$50^{50}<49^{51}$$
$$\left(\frac{50}{49}\right)^{50}<49$$
$$\left(1+\frac{1}{49}\right)^{50}<49$$
$$\left(1+\frac{1}{49}\right)^{50} \approx e \approx 3$$
$$3<49$$
Therefore we conclude that $50^{50}<49^{51}$
A: Note that $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\dots+x^n$
$50^{50}$ and $49^{51}$
$50^{50}$ and $49^{50}\cdot49$
$(\frac{50}{49})^{50}$ and $49$
$(1+\frac{1}{49})^{50}$ and $49$
$1+50(\frac{1}{49})+\underset{\text{Negligible terms}}{\underbrace{\frac{50\times49}{2}(\frac{1}{49})^2+\frac{50\times49\times48}{6}(\frac{1}{49})^3+\dots+(\frac{1}{49})^{50}}}$ and $49$
$1+\frac{50}{49}$ and $49$
Clearly, $1+\frac{50}{49}<49$. Hence $50^{50}<49^{51}$. Thus, B is the correct choice.
