# 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

(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?

Hint: Use the inequality $$\left(1-\frac{1}{x}\right)^x>\frac{1}{x-1}$$

• From where did you come with this equality? How can we derive it? It holds true for $x<0$ or $x>4.14...$. – Hussain-Alqatari Jul 7 '19 at 13:51
• @Hussain-Alqatari We only need it for large enough positive $x$, since the left-hand side increases while the right-hand side decreases. For $x>5$, $\left(1-1/x\right)^x>0.32768$ while $1/(x-1)<0.25$. – J.G. Jul 7 '19 at 13:57
• Lets say $$x=50$$ so we have to show that $$(x-1)^{x+1}>x^x$$ this is the inequality above. – Dr. Sonnhard Graubner Jul 7 '19 at 13:58

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 $$\left(1-\frac{1}{50}\right)^{50} = \left(\left(1-\frac{1}{50}\right)^{25} \right)^2 \ge \left(1-\frac{1}{50}\cdot 25\right)^2 = \left(\frac12\right)^2 = \frac14 > \frac{1}{49}$$

Use Bernoulli's inequality with $$x=-\frac{1}{50}$$ and $$r=48$$:

$$\frac{49^{51}}{50^{50}}=\frac{49^3}{50^2}\cdot\frac{49^{48}}{50^{48}}=\frac{49^3}{50^2}\left(1-\frac{1}{50}\right)^{48}\ge\frac{49^3}{50^2}\left(1-\frac{48}{50}\right)=\frac{2\cdot49^3}{50^3}>1$$

• One more application of Bernoulli to show that $\frac{2\cdot49^3}{50^3}\ge\frac{2\cdot47}{50}\gt1$ might help those who can't compute $49^3$ in their heads (although $50^3$ is quite possible). – robjohn Jul 11 '19 at 15:06

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 $$(1)$$.

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 $$50^{50}<49^{51}$$

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.