Which is greater?

We can use binomial expansion to help compare these two numbers:

$\begin{aligned} 71^{70} & = (70 +1)^{70} \\ & = 70^{70} + 70(70^{69}) + \frac{70\dot{}69}{2}(70^{68}) + \frac{70\dot{}69\dot{}68}{2\dot{}3}(70^{67}) +... + 1 \quad \small \color{#3D99F6}{[71 \text{ terms}]} \\ & = 70^{70} + 70^{70} + \frac{69}{2}(70^{69}) + \frac{69\dot{}68}{2\dot{}3}(70^{68}) +... + 1 \\ & < 70(70^{70} ) = 70^{71} \end{aligned}$

$\Rightarrow \boxed{70^{71}} > 71^{70}$

$71^{70} \boxed{} 70^{71}$

$\dfrac{\ln (71)}{71} \boxed{} \dfrac{\ln(70)}{70}$

Consider function $f(x)=\dfrac{\ln (x)}{x}$ and the derivative is $\dfrac{1-\ln(x)}{x^{2}}$ since we have this we know that when $x> e$ it's a decreasing function.

Hence, when the x value increase the value of function decrease.

So,

$\dfrac{\ln (71)}{71} \boxed{<} \dfrac{\ln(70)}{70}$

then

$71^{70} \boxed{<} 70^{71}$

Suppose $71^{70} > 70^{71} \Rightarrow 71^{70} > 70 \cdot 70^{70}$

We can rewrite this inequation as

$\left(\frac{71}{70}\right)^{70} > 70$

Or...

$\left(1+\frac{1}{70}\right)^{70} > 70$

The expression on the left is now on the form

$\left(1+\frac{1}{x} \right)^x$

Which we know is never greater than the euler constant (2.71828) for increasing values of x, and thus can't be greater than 70. So it turns out then that the opposite is true ( $70^{71} > 71^{70}$ )

We can solve this using a Taylor Expansion. First we consider the following Taylor Expansion of $( 1-x)^{n} \approx 1-nx$ , where $x \ll 1$ . It follows then that

$(71)^{70} = (70 +1)^{70 }= 70^{70} (1+\frac{1}{70})^{70} \approx 70^{70}(1 + 70\frac{1}{70}) = 70^{70} \times 2$

Concerning the other term,

$70^{71} = 70^{70} \times 70^{1} = 70^{70}\times 70$ .

Thus,

$71^{70} < 70^{71}$ .

Solved like a true Physicist.

Let's say 71 = 70*X

71^70 = 70^70 * X^70

Multiply with 70 on both side

71^70 * 70 = 70^71 * X^70

71^70 / 70^71 = X^70 / 70

Now if, X^70/ 70 >1 then 71^70 is more else 70^71 is more

Approx Calculation for X is 1.01. So X^70 will turn out to be approx = 1 . Hence X^70/ 70 = 1 / 70 (approx.) < 1

So 70^71 is more :). Hope this was a simple solution and easily to explain to others.

This how I solve the majority of problems: take a small sample data: $2^{3}= 8$ $3^{2}=9$ $4^{3} = 64$ $3^{4} = 81$ $5^{4} = 625$ $4^{5} = 1024$ By induction we can reason that for n > 2 $n^{n+1} > (n+1)^{n}$
$Therefore: \boxed{70^{71} > 71^{70} }$

Here's a quick three-step method that works for problems of the same type with sufficiently large numbers.

1. Consider the ratio $\frac{71^{70}}{70^{71}}$ .

2. Rewrite the ratio as $\frac{1}{70} . (1 + \frac{1}{70})^{70}$ .

3. Recognize e (~ 2.71828) is an upper-bound of $(1 +\frac{1}{70})^{70}$ , since $e = \lim_{n\to\infty} (1 + \frac{1}{n})^{n}$ and the sequence is monotonically increasing.

It now becomes clear that $\frac{71^{70}}{70^{71}} < \frac{e}{70} < 1$ , implying $71^{70} < 70^{71}$ . QED.

divide both side by 7^70

(71/70)^70 vs 7

It helps to look at the general case.

(n+1)^n < n^(n+1) for all n>~2.29317, which is the solution to the equation (n+1)^n = n^(n+1)

Here n = 70, which is much larger than the value of ~2.29317

1. 71^{70} = (70+1)^{70}
2. 70^{71} = (70)^{(70+1)}

From 2, 70^{70}x70^{1}, therefore will be greater.

Please correct me if i''m wrong :)

"1^t = 1 (t € R)" 71/70 ~= 1 (71/70)^70 = a small number (71/70)^70<70 of course 71^70<70^70 * 70 71^70<70^71

take log on both sides. 71 log 70 and 70 log 71 now log x < x now, 71 > 70 log 70 < log 71 but multiplied together 71 log 70 > 70 log 71 since log x has smaller effect than x .

log 71 > (log 71 - log 70) x 71, so 70 ^ 71 is greater

Is this a valid calculation?

Of course, the easiest solution is to plug both directly in calculator. You will get difference of 9.6579e130 approximately.

However, it can be proven, that $\forall x, a, where x > a > e, a^{x} > x^{a}$

Therefor $70^{71}>71^{70}$

Let's just say this: Powers and exponentials get big. The power and base are not negative, or 1, but in fact REALLY BIG. This means the power can easily take over, even if the base is one less! Use your intuition with this one( or, if you REALLY want, your calculator)

We know that $a^{a+1}$ & $(a+1)^a$ that $a^{a+1}>(a+1)^a$ such that $a$ is a integer & $a>2$ so $70^{71}>71^{70}$

Write the fraction $\frac{71^{70}}{70^{71}}$

then scompose $70^{71} = 70^{70}*70$

then we collect $(\frac{71}{70})^{70}*\frac{1}{70}$

noticed that $\frac{71}{70} \approx 1.00$

so $\frac{71^{70}}{70^{71}} = (\frac{71}{70})^{70}*\frac{1}{70} \approx \frac{1}{70} < 1$ .

If we have a fraction less than one the denominator is bigger than the numerator so the answer is ${70^{71}}$

From the following evaluations it can be infered that ${\rm{7}}{{\rm{0}}^{71}} > {71^{70}}$ as the values on the left of the inequality sign are increasing faster than those on the right of the inequality sign \begin{array}{*{20}{l}} {{0^1} < {1^0}}&{{\text{as }}0 < 1} \\ {{1^2} < {2^1}}&{{\text{as }}1 < 2} \\ {{2^3} < {3^2}}&{{\text{as }}8 < 9} \\ {{3^4} > {4^3}}&{{\text{as }}81 > 64} \\ {{4^5} > {5^4}}&{{\text{as }}1\,024 > 625} \\ {{5^6} > {6^5}}&{{\text{as 1}}5\,625 > 7\,776} \\ {{6^7} > {7^6}}&{{\text{as }}279\,936 > 117\,649} \end{array}

Take small numbers do calculations e.g. 3 power 4 is equal to 81 and 4 power 3 is equal to 64 This way small number with greater power comes out to be larger number so 70 base power 71 is larger than 71base power 70.

$\frac{70ln71}{71ln70}$ = 0.989

meaning the denominator has a higher value. thus 70^71 has the larger value

It can be easily derived using induction

Clearly 70^71 is greated than 71^70 as 70 is to a larger power.

We know that if a<b de have ln(a)<ln(b), So, ln(71^70)=70ln(71)=70ln(70) + 70ln(71/70)

And ln(71/70) is very close to zero On the other hand de have ln(70^71)=71ln(70)=70ln(70)+ln(70)

Then ln(70)>70ln(71/70) So, ln(70^71)>ln(71^70) And, 70^71>71^70

Well.... 2^8 is greater than 8^2....therefore 70^71 is greater than 71^70

Let's try to solve this using calculus.

$\large{71^{70} \boxed{?} 70^{71}}$

$\large{71^{\frac{1}{71}} \boxed{?} 70^{\frac{1}{70}}}$

Now, let's look at the function $\large{n^\frac{1}{n}}$ .

Let's differentiate it.

$\large{\frac{d}{dn} n^\frac{1}{n} = \frac{d}{dn} e^{ln\left({n^\frac{1}{n}}\right)} = \frac{d}{dn} e^\frac{ln\left(n\right)}{n} = \left(\frac{d}{dn} \frac{ln\left(n\right)}{n} \right)e^\frac{ln\left(n\right)}{n} }$ $\large{= \frac{1 - ln\left(n\right)}{n^2}e^\frac{ln\left(n\right)}{n} = ln\left(\frac{e}{n}\right)n^{\frac{1}{n} - 2} }$

Which is negative iff $n > e$ .

This means that $n^\frac{1}{n}$ is strictly decreasing when $n > e$ which implies that $\large{71^{\frac{1}{71}} < 70^{\frac{1}{70}}}$ which in order implies that $\large{71^{70} < 70^{71}}$ .

Thus, $\boxed{70^{71}}$ is greater of these 2 numbers.

(This also proves that $n^m < m^n$ , if $e < m < n$ )

I don't know what you guys are thinking but I just thought about what calculators did when there were too many zeros

71^70 = X so 70log71=log X hence 10^129.588 =X
70^71 =Y so 71log70= log y hence 10^131.0019608 =Y

Y>X

(1+1/n)^n-->e when n is toward infinity provided n is a positive integer, it will always smaller than e (71/70)^70<e<70 (71^70)/(70^70)<70 71^70<70^71

71^70=(70+1)^70=70^70+141

70^71=70^70 70=70^70+(70^70) 69

(70^70)69>141

70^71>71^70

If we have $y^{x}$ and $x^{y}$ when $x > y$

Then $y^{x}$ > $x^{y}$

Only if {x,y} $\neq 1$

And {x, y} $\in R$

But when:

$x = 3$ $\vee$ $y = 2$ $\to$ $y^{x}$ < $x^{y}$

$x = 4$ $\vee$ $y = 2$ $\to$ $y^{x}$ = $x^{y}$

Is it bad that I did this in my head and have no logic behind it? In my head 70^71 > 71^70 is the same as 1 + 1 = 2. Sure I could write some crazy theorem to prove it or I could just know it is true and leave it at that.

70^71=[70^(71/70)]^70=(74.38...)^70 > 71^70

I figured I would post my reply to a solution that I saw in this thread. It was basically a correct idea using intuition about growth rates, but it wasn't 100% spelled out.

Let <> stand for the unknown inequality, and >< its opposite if needed.

71^70 <> 70^71

Well, we know that 70^70 = 70^70, and we increase one number on each side by one. Which grows more?

The question is (a+E)^a <> a^(a+E).

a = 70, E = 1. The important features are that a > e = 2.718... , and E>0.

Log helps. (You can use log 10, but the "pure math" convention is log e, sometimes written as ln)

The question, then, is a log(a+E) <> (a+E) log(a)

Equality holds if E=0, but for E>0, both sides are positive. Treating E as the only variable, the LHS has derivative (that is, d/dE) equal to a/(a +E), and the RHS has derivative log a. So as E increases, the LHS increases more slowly. In fact, the RHS grows more than (log a) times as fast! log a > 1 > a/(a+E), since a > e and E>0.

a log (a + E) < (a + E) log a, and so 70 log(71) < 71 log(70)

You can be more accurate, too. The LHS could have grown (log a) times as fast and still lost the growth race, because a/(a+E) < 1. We could prove that:

a log (a + E log a) < (a + E) log a.

So we even have

70 log (70 + log70) < 71 log 70, (Calculator says: 301.519... < 301.643...)

or

(70 + log 70) ^ 70 < 70 ^ 71. (The LHS is bigger than 71^70) (big numbers...)

In other words, (since log(70) = 4.248... )

(74.248...)^70 < 70^71

just follow this 2^3 < 3^2

Yah we can do by binomial theorem too. But my friend have already done that! So I have a different approach. Let 71 =2, and 70 =2 so we have 2^70and 2^71 which is bigger we know! Just fun!