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55

In the usual computer science jargon, exponentiation is right-associative, which means that $x^{y^z}$ should be read as $x^{(y^z)}$, not $(x^y)^z$. One way to remember this is that $(x^y)^z = x^{yz}$, so it would be silly if out of the two possibilities, $x^{y^z}$ meant the one that can be expressed without using two tiers of superscripts.

17

Usually, a^b^c is taken to mean a^(b^c). This is purely an issue of the definition of notation so deep "why" answers aren't super likely. The main thing is that we have the identity (for positive $a$): $$(a^b)^c=a^{bc}$$ so it would make little sense to make that the default order, given that it reduces to a simpler form, whereas $a^{(b^c)}$ doesn't reduce. ...

16

Here is your "proof" presented differently: We have $e^{i\pi}=-1=\frac{1}{-1}=\frac{1}{e^{i\pi}}=e^{-i\pi}$. So far everything is right. Now our idea is to take both sides to the power of $i$: $(e^{i\pi})^i=(e^{-i\pi})^i$. The erroneous conclusion would appear if you used the identity $(a^b)^c=a^{bc}$. And here lies the problem: this identity doesn't hold ...

16

Let $x = \mathrm{e}$ and $y = \ln(2)$, then $x^y = \mathrm{e}^{\ln(2)} = 2$.

15

Let $x=3^{1/2}$ and $y=\log_{3}(4)$. Then $x^y=2$. The proof that $x$ is irrational is familiar. For $y$, suppose $y=p/q$ where $p$ and $q$ are positive integers. Then $3^{p/q}=4$, so $3^p=4^q$. This is impossible, since $4^q$ is even and $3^p$ is odd.

13

By continuity of the function $x^x$, there is an $a$ such that $$a^a=2.$$ This number is irrational. Otherwise, let $a$ be the irreducible fraction $p/q$ and $$\left(\frac pq\right)^{p/q}=2,$$ is equivalent to $$p^p=2^qq^p,$$ which implies that $p$ is even and $q$ is even, a contradiction. By the way, as $$\ln(a^a)=a\ln(a)=\ln(a)e^{\ln(a)},$$ we ...

12

You're not doing anything wrong, per se; the issue is that in order to define square root as a single-valued function on $\mathbb{R}_{\geq 0}$, we (somewhat arbitrarily) choose the positive value. So both $-1$ and $1$ square to $1$, but the square root of $1$ is only $1$, not $-1$. This becomes a bit more transparent if we replace $14$ by $2$: $$... 10 You have put your finger precisely on the statement that is incorrect. There are two competing conventions with regard to rational exponents. The first convention is to define the symbol a^x for a > 0 only. The symbol \sqrt[n]{a} is defined for negative values of a so long as n is odd, but according to this convention, one wouldn't write ... 9 In real numbers, the standard definition of rational exponents only permits fully reduced fractions in the exponent. Example definition from Sullivan's College Algebra: Definition. If a is a real number and m and n are integers containing no common factors, with n \geq 2, then$$a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]a\right)^m$$provided that ... 8 The notion that (a^b)^c=a^{bc} has to be abandoned in complex analysis. Or, you have to allow that a^b is a multi-valued function and then you can actually say that (one of) the values of 1^x is \cos(2\pi x)+i\sin(2\pi x). With multi-valued functions you can say "All of the values of a^{bc} are values of (a^b)^c," but not visa versa. ... 7 Not a rigorous proof at all, but you could take the xeth root of both sides, and then we have$$ e^{1/e} \geq x^{1/x} That x^{1/x} attains a maximum at x = e can be shown fairly straightforwardly. ETA2: A picture is worth—well, a lot of words, if not quite a thousand: ETA: OK, a discussion of why x^{1/x} attains a maximum at x = e. To ... 7 Take the natural logarithms of both sides, then $$\ln 3 (\ln 3 + \ln x)=\ln 4 ( \ln 4 + \ln x)$$ Thus $$\ln x =\frac{(\ln 3)^2 - (\ln 4)^2}{\ln 4 - \ln 3}=-(\ln 4+\ln 3)=-\ln 12=\ln \frac{1}{12}.$$ Since \ln x is injective, x=\frac{1}{12}. 6 Hint: you can establish the right order between 5^{19} and 2^{39} as follows \begin{align} 5^{19} &= 5^{20-1}\\ &=\frac{5^{20}}{5}\\ &=\frac{(5^2)^{10}}{5}\\ &=\frac{25^{10}}{5}\\ &>\frac{16^{10}}{2}\\ &=\frac{(2^4)^{10}}{2}\\ &=2^{39} \end{align} then note that \begin{align} 52^7 &= 52^{10-3}\\ ... 5 First, we can easily compute some small powers manually to see the equalities in\phantom{(\ast)} \qquad 2^{11} = 2048 < 3^7 = 2187 < 5^5 = 3125. \qquad (\ast)$$Multiplying both sides of the first inequality in (\ast) by 2^{28} = 16^7 gives the left-hand inequality in$$2^{39} < 3^7 \cdot 16^7 = 48^7 < 52^7 .$$On the other hand, ... 5 Hint: multiply by$$\frac{\sqrt{4^n+3^n}+2^n}{\sqrt{4^n+3^n}+2^n}$$Then divide both numerator and denominator by 3^n. Keep in mind that 3^n=\sqrt{3^{2n}}=\sqrt{9^n}. 5 Let f:\mathbb{R}^+\to\mathbb{R} given by f(x)=e^x-x^e. Then f^\prime(x)=e^x-ex^{e-1}=e(e^{x-1}-x^{e-1}) and f^{\prime\prime}(x)=e(e^{x-1}-(e-1)x^{e-2})=e(e^{x-1}-ex^{e-2}+x^{e-2}). Thus f^{\prime}(e)=0 and f^{\prime\prime}(e)=e^{e-1}>0. So f has a local minimim in e. Since \lim_{x\to\infty}f(x)=\infty, thus f attains his absolut ... 5 This is about a simplified as you can get. You can also write it as 5^k(5^k+1). if you prefer that. There's no simple formula like a^k. 5 In the denominator (2n)!= (2n)(2n-1) \dots (n+1)\cdot n!. Each of the factors from (n+1) to (2n) are larger than n; there are n of these factors. So you can show that this sequence is less than 1/n!. 4 The root of the problem is that \sqrt{ 1\;}=+1, unambiguously by definition, but 1^\frac12=(e^{2\pi i k})^{\frac12}=e^{\pi i k}\;\;\forall k\in\mathbb Z is not, it results to \pm1 depending on k being odd or even. Taking roots, you have to choose a branch, like the commonly accepted branch that \sqrt x \ge 0 for x\ge0. We could just as well have ... 4 The notation helps here; the exponent (which is the part that's raised) always acts like it has parentheses around it. So x^{y^z} means x^{(y^z)}. Similarly, x^{y+z} means x^{(y+z)} and x^{yz} means x^{(yz)}, even though exponentiation has higher precedence than addition or multiplication (so x+y^z means x+(y^z) and xy^z means x(y^z)). 4 The reasons is that the exponential function(something^x) grows much faster than the power function (x^{something}) So in general you would expect that for every a, b > 1something like this hold$$a^x \ge x^b$$at least for x big enough Now I don't think it's particularly meaningful that for the special case a = b = e the inequality holds ... 4 In complex numbers exponentiation rules are a bit different, in this case$$(e^{2 \pi i})^x\not\equiv e^{2 \pi i x}$$4 As a hint, I'd suggest: Write out (a^{\frac{1}{2}}\times b^{\frac{1}{3}})^3, Factorize 432 into prime numbers. 4 Note that$$432 = 2^4\cdot 3^3 = 3^3\cdot4^2$$and the LHS can be written as$$(a^\frac{1}{2} \times b^\frac{1}{3})^6 = a^3\cdot b^2$$So a = 3, b = 4 \Rightarrow ab = 12. 4 The issue is that a^{\frac{1}{n}} is multivalued. You could arguably simplify the first calculation into 1 = \sqrt{1} = -1. Taking different branch cuts is how the "paradox" arises. Essentially, in the context of the reals (or even the complex numbers) \sqrt{a} is one name for two functions, say \sqrt[+]{a^2} = a and \sqrt[-]{a^2} = -a. All the ... 4 Consider the more general case: |f(x)-f(y)|\le |x-y|^{1+\varepsilon}, with \varepsilon>0. Then f is differentiable with f'=0 because$$ \lim_{x\to x_0} \left|\frac{f(x)-f(x_0)}{x-x_0}\right| \le \lim_{x\to x_0}|x-x_0|^{\varepsilon} = 0 $$So, \sqrt2 is a red herring. Its only relevant property is \sqrt2>1. 4 Let x be the number we are after. Then 11x\equiv 11^{112}\equiv 1\pmod{113}. So we are looking for the modular inverse of 11. Multiply by 11 and reduce mod 113. We get 8x\equiv 11\pmod{113}. This is equivalent to 8x\equiv 124, which is equivalent to 2x\equiv 31, which is equivalent to 2x\equiv 144, which is equivalent to x\equiv ... 4 To compare 2^{39} and 5^{19} we have$$2^{39} = 2 \cdot 2^{38} = 2\cdot 4^{19} = 2 \cdot 4^4 \cdot 4^{15} = 512 \cdot4^{15} < 625 \cdot 5^{15} = 5^{4}\cdot 5^{15} = 5^{19}$$. 3 Rewrite the expression as$$ 2^n \sqrt{1 + (3/4)^n} - 2^n = 2^n\left(\sqrt{1 + (3/4)^n} - 1\right) = \frac{\sqrt{1 + \epsilon^n} - 1}{\delta^{n}}  with $\epsilon = 3/4, \, \delta = 1/2$. Now use L'Hopital's Theorem. Since $\epsilon > \delta$, you will find that the limit is $\infty$.

3

If $a^a=2$, then $a$ is irrational: If $a=p/q$, then $(p/q)^p=2^q$ is an integer, so $p/q$ is an integer.

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