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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 ...


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 $xe$th 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 \begin{equation} \ln 3 (\ln 3 + \ln x)=\ln 4 ( \ln 4 + \ln x) \end{equation} Thus \begin{equation} \ln x =\frac{(\ln 3)^2 - (\ln 4)^2}{\ln 4 - \ln 3}=-(\ln 4+\ln 3)=-\ln 12=\ln \frac{1}{12}. \end{equation} 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 $a, b > 1$something 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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