# Tag Info

54

We just have to show that $a^{a-b} \ge b^{a-b}$. This is equivalent to $(\frac{a}{b})^{a-b} \ge 1$. If $a \ge b$, then $\frac{a}{b} \ge 1$, Also $a-b \ge 0$. A number greater than $1$ raised to a positive exponent is clearly greater than $1$. If $a \le b$, then $\frac{a}{b}\leq 1$. $a-b\leq 0$. A positive number less than $1$ raised to a negative ...

43

Raise them both to the power of $6$. Since they are both positive, their order will be preserved and you will get: $$\left({\dfrac{1}{2}}\right)^2=\frac{1}{4} > \frac{1}{27}=\left({\dfrac{1}{3}}\right)^3$$

39

No need to do any calculations at all: since we are talking about numbers between $0$ and $1$, a cube root is larger than a square root: $$\Bigl(\frac12\Bigr)^{1/3}>\Bigl(\frac12\Bigr)^{1/2}>\Bigl(\frac13\Bigr)^{1/2}\ .$$

25

$$\log(a^a b^b)=a \log a + b \log b$$ $$\log(a^b b^a)=a \log b + b \log a$$ Thus, by the rearrangement inequality, because $\log$ is strictly increasing, $$\log(a^a b^b)\geq \log(a^b b^a)$$ Similarly, because $\log$ is strictly increasing, $$a^a b^b \geq a^b b^a.$$

14

For any $x \in \mathbb{R}$, $2^x > 0$ and $4^x > 0$, therefore $$2^x + 4^x + 12 > 0 + 0 + 12 = 12 > 0$$ Therefore there is no real solution.

12

My favorite paper about $x^x$ is The $x^x$ Spindle, which appeared in Mathematics Magazine back in 1996. The main idea is to visualize the fact that we can write it as $$x^x = e^{x (\ln(x)+2k\pi i)}.$$ Note that for each choice of $k$, we get a different branch of the logarithm. Given any real number $x$, most of these branches will be complex valued. ...

10

$$n^{\frac{1}{n}} \leq (n+9)^{\frac{1}{n}} \leq (2n)^\frac{1}{n}$$ For $n$ greater than, say, $9$. Apply "squeeze theorem"

8

$$\left(\frac ab\right)^{a-b}-1=\frac{a^ab^b-a^bb^a}{b^aa^b}$$ If $a=b, \left(\dfrac ab\right)^{a-b}=1$ Else if $a>b;\dfrac ab>1$ and $a-b>0\implies \left(\dfrac ab\right)^{a-b}>1$ Similarly if $a<b$

6

There are a few things going on. Jump to the end for discussion of complex numbers and what exactly WolframAlpha is plotting. When you plot the function, WolframAlpha and most plotting systems don't restrict themselves to special inputs like "only rational numbers with odd denominator." Even if you try using an odd denominator, most computational software ...

5

When is $x^y > y^x$ ? When $x^{1/x} > y^{1/y}$. Let's look at the function $x^{1/x}$. Differentiating, we find it has a maximum at $x=e$. Since $1/2$ and $1/3$ are both less than $e$, the one that's nearer wins. So $(1/2)^2 > (1/3)^3$, so $(1/2)^{1/3} > (1/3)^{1/2}$. But more to the point, this shows that $e^\pi > \pi^e$, which might be a lot ...

5

Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$. $b$ can be expressed as $a+x$, where $x$ is some positive integer. to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$ Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$ $= a^a (a+x)^{a+x} - a^{a+x} (a+x)^a$ ...

5

This inequality is equivalent to $a\ln a+b\ln b\geq a\ln b+b\ln a$, which is obvious by Rearrangement or it's $(a-b)(\ln a-\ln b)\geq$, which is a proof of Rearrangement.

5

let $u = 2^x$, then $4^x = (2^2)^x = 2^{2x} = 2^{2x} = (2^x)^2 = u^2$. Thus, $2^x + 4^x + 12 = 0$ becomes, $$u + u^2 + 12 = 0$$ Using the quadratic equation will solve $u$, which is really $2^x$. To solve for the $x$, just take $\log$ on both sides of the solution, then after rearranging, you should be able to solve for $x$.

5

You could try the binomial expansion of $(1+0.05)^{10}$ and stop calculating terms after they become small enough to not affect your required degree of accuracy

4

Of the several different but related definitions for exponentiation, the one that accepts a rational exponent should be phrased to exclude negative bases, in order to avoid exactly this problem. In fact, the only one of the usually encountered definitions that give meaning to a negative integer to a non-integral power is the one for complex numbers, which ...

4

If you don't want to depend on the "trick" of raising to the sixth power, you can compare the logs: $\frac 13 \log \frac 12=\frac {- \log 2}3$ and $\frac 12 \log \frac 13=\frac {-\log 3}2$ Now $\frac 12 \gt \frac 13$ and $\log 3 \gt \log 2$, so $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} ... 4 The following is a variation of the visualization of the function$x^xthat I described in this answer. It's not clear to me how to explain it to middle school kids, though. Specifically, if you want to explain the WolframAlpha output to middle schoolers, then they've got to know that $$(-2)^x = e^{x\log(-2)} = e^{x(\log(2) + i\pi)} = 2^x e^{xi\pi} = ... 3 ((\frac{1}{2})^{\frac{1}{3}})^6=(\frac{1}{2})^2=\frac{1}{4} ((\frac{1}{3})^{\frac{1}{2}})^6=(\frac{1}{3})^3=\frac{1}{27} So as it is obvious from the above relations, ((\frac{1}{2})^{\frac{1}{3}})^6>((\frac{1}{3})^{\frac{1}{2}})^6, so we can say (\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}} 3$$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$Now it is obvious that$$\sqrt[3]2<\sqrt3$$Thus$$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$3 Let's write$$ \begin{align} &\log\log n = L_2(n), \\ &\log\log\log n = L_3(n), \\ &\log\log\log\log n = L_4(n). \end{align} $$Then from x^{x^x} = n we get$$ x\log x + L_2(x) = L_2(n), \tag{1} $$and so, since x \to \infty as n \to \infty, we have$$ x\log x \sim L_2(n) \tag{2} $$as n \to \infty. Taking logs of this yields$$ ... 3 $$a^a \ b^b \;?\; a^b \ b^a \\ \frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\ \left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\ \left(\frac{a}{b}\right)^{a-b} \;?\; 1$$ ifa \ge b$, then$c = \frac{a}{b} \ge 1$, and$d = a-b \ge 0$. Thus$c^d \ge 1$, so$?$is$\ge$. if$a < b\$, then: $$\left(\frac{a}{b}\right)^{a-b} \\ = ... 3 Hint: Suppose that \sqrt{1}+\sqrt{2} is of the shape c^r, where c is a positive rational and r=\frac{m}{n} where m and n are integers, with n\gt 0. Show by taking the n-th power of both sides that this implies that \sqrt{2} is rational. 3 You have 2 terms of 2^{n+1}, meaning you have$$2\cdot 2^{n+1} -1 = 2^{(n+1)+1} - 1 = 2^{n+2}-1$$3 By the difference of perfect squares.$$\large(a+b)(a-b) = (a+b)a - (a+b)b = a^2+ab-ab-b^2=a^2-b^2$$We just have to let a = 2^6 and b=2^3. 2 See OEIS sequence A074981 and references there. 10 does have a solution as 13^3-3^7, but apparently no solutions are known for 6 and 14. 2 I don't have an answer to your question, but I did search through quite a few set theory books this morning and I made notes of what I found in case you or others are interested. The topic seems less covered in books than I expected, and I suspect you'll have to consult journal articles to find much of significance (unless you can read Hessenberg's and ... 2 Note that$$2^{n+1}+2^{n+1}=2\cdot 2^{n+1}=2^1\cdot 2^{n+1}=2^{n+1+1}=2^{n+2}.$$2 Take \log:$$ \log L = \lim_{n\to\infty}\log(n+9)^{\frac{1}{n}} = \lim_{n\to\infty}\frac{\log(n+9)}n = \cdots $$2 Let$$f(t)=\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t} =\{g(t)\}^{1/t}$$and if \lim_{t\to 0}f(t)=L, then$$\begin{aligned}\log L &= \log\left\{\lim_{t \to 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\right\}\\ &= \lim_{t \to 0}\,\log\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\text{ (by continuity of log)}\\ &= \lim_{t \to ...

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