# Tag Info

10

\begin{align} \frac{x^x}{(2x)!} & = \frac{\overbrace{x\cdots x}^{x\text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots x}_{x\text{ factors}} \cdot \underbrace{(x+1) \cdot (x+2) \cdots (2x)}_{x\text{ factors}}} \\[10pt] & = \frac 1 {x!} \cdot \underbrace{\frac x {x+1} \cdot \frac x {x+2} \cdot \frac x {x+3} \cdots \frac x {x+x}}_\text{This is $<1$.} ...

9

A number $n \in \mathbf N^+$ has $\lfloor\log_{10}n\rfloor+1$ digits. For $n = 5^{4^{3^2}}$, we have \begin{align*} \log_{10} 5^{4^{3^2}} &= 4^9 \cdot \log_{10} 5\\ &= 262\,144 \cdot \log_{10} 5\\ &\approx 183\,230.8 \end{align*} So $n$ has $183\,231$ digits.

8

$$\left\lfloor \frac{1763.192674118048}{\log 10} \right\rfloor = 765$$ This is the logarithm base $e,$ so $\log 10 \approx 2.30258509$ Since $$\frac{1763.192674118048}{\log 10} \approx 765.7448488$$ we find that your number is $$e^{1763.192674118048} \approx 5.5571 \cdot 10^{765}$$ because $$10^{0.7448488} \approx 5.5571$$

7

In the equation $x^3+\frac{1}{x^3}=18$, multiply everything by $x^3$ to get $x^6-18x^3+1=0$. Then let $y=x^3$ and solve $y^2-18y+1=0$, then substitute back in to get $x$. Then you can input the value into $x^{11}+\frac{1}{x^{11}}$ directly. It winds up being $x^{11}+\frac{1}{x^{11}}=39603$ (regardless of whether you use the plus or the minus in the ...

7

Note that $$\frac{n^n}{(2n)!}=\frac{\overbrace{n\cdot n\cdot\ldots \cdot n}^{n \text{ factors}}}{n!\cdot\underbrace{(n+1)(n+2)\ldots(2n)}_{n\text{ factors}}}\le \frac1{n!}$$

6

$$A^n=\begin{pmatrix}1&nb\\0&1\end{pmatrix}$$ Suppose this is true until $n$. Then $$A^{n+1}=A.A^n$$ Computing the right side almost completes the proof.

5

They can be, but they don't have to be. One of the simplest setups is to define $\exp$ to be the unique function satisfying $\exp'=\exp$ and $\exp(0)=1$; then define $\ln$ to be the inverse of $\exp$; then define $a^x=\exp(x \ln(a))$ for $a>0$. Then the exponent and logarithm rules are theorems. Most of the proof of the rules is just elementary algebra; ...

5

It's transcendental by the Gelfond-Schneider theorem.

5

We may also avoid De l'Hopital theorem. Since $$\lim_{t\to 0}\frac{e^t-1}{t}=1\quad\text{and}\quad \lim_{x\to 0^+} x\log(x)=0,$$ we have: $$\lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$

5

Looks like it's impossible for two distinct numbers to be "power equivalent". Really, $a_{n+1}=a^{b_n}$ and $b_{n+1}=b^{a_n}$. Now, if we use the limit definition and choose $n$ large enough, $a_{n+1}$ is close to $b_{n+1}$, which means for $a_n$ that it is close to ${\log a\over\log b}b_n$, but at the same time it is close to $b_n$, which is impossible ...

5

$$(5^{2015})(2^{2018}) = (5^{2015})(2^{2015})(2^3) = (5\cdot 2)^{2015}2^3$$ $$= 10^{2015}2^3 = 8\cdot 10^{2015}$$ Notice that this is just the digit $8$ followed by $2015$ zeroes, so the sum is just $8$

4

Use L'Hopital's rule to get: $$\lim_{x \to 0} x^x(\ln x+1)=\lim_{x \to 0}x^x \ln x-\lim_{x \to 0} x^x=(\lim_{x \to 0} x^x \ln x) -1=((\lim_{x \to 0} x^x)(\lim_{x \to 0} \ln x))-1=(1)(-\infty)-1=-\infty$$ The limit does not exist.

4

Without induction: Write $A$ as $I+bE$, where $I$ is the unit matrix of rank $2$ and $E=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. As both matrices commute with each other, we can apply the binomial formula in the ring $M_2(\mathbf R)$, noting that $E^2=0$: $$A^n=\sum_{k=0}^n\binom nk I^{n-k}b^kE^k=I+nbE=\begin{bmatrix}1&nb\\0&1\end{bmatrix}.$$

3

Hint: $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)=18.$$ Let $$x+\frac{1}{x}=t.$$ Then your equation reduces to $(t(t^2-3))=18$. Solve this cubic and you get $x+\frac{1}{x}.$ Now I think you can carry on after that.

3

A little deus ex machina but this seems to be related to the golden ratio $\phi = \frac{1+\sqrt 5}{2}$, which is the solution of $x+\frac1x = 1$, and to the Fibonacci sequence $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657, \ldots$ In particular $x^3+\frac1{x^3} = 18$ has the solutions $x^3=5+8\phi$ and its ...

3

By the Catalan conjecture (https://en.wikipedia.org wiki/Catalan's_conjecture ) solved by Mihailescu in 2002, 8 and 9 are the only consecutive powers, so yes, your question can be answered.

3

As $m^\frac{a}{b}=n$ therefore, $(m^\frac{a}{b})^b=(n)^b$ $m^a=n^b$

3

$\newcommand{\Reals}{\mathbf{R}}$If $n$ is a positive real number and $f_{n}:\Reals^{3} \to \Reals$ is defined by $$f_{n}(x, y, z) = (x^{2} + y^{2} + z^{2})^{n/2}, \tag{1}$$ then the level sets of $f_{n}$ are spheres centered at the origin, independently of $n$. (The level set at "height" $0$ consists of the origin alone, but may be viewed as a sphere of ...

3

The question is really "how many subsets of $\{1,3,5,7,11,17\}$ are there, not counting the null set?" As you said each element is either in the subset or it isn't. Thus each element has two choices, so there are $2\cdot2\cdot2\cdot2\cdot2\cdot2=2^6$ subsets, and we subtract one more since we also just counted the null set, which we don't want. This may be ...

2

Let $p =$ number of elements of a set $S$ Maximum number of subsets of $S$ is $2^p$ which includes a null set too. Since, we want only odd numbers so we take a set of odd numbers O ={1,3,5,7,11,17} which has cardinality $6$. Then the maximum number of sets that can be formed from this set is $2^6 -1=63$ because we have to discard the null set.

2

Assuming $x$ is an integer in your question (so that I'll use $n$ instead of $x$, for the sake of my own ease of mind): a simple way, which uses a big (and quite overkill) "hammer," is to invoke Stirling's approximation: $$(2n)! \operatorname*{\sim}_{n\to\infty} 2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}}$$ and look at the limit (when $n\to\infty$) of $$... 2$$m^{\frac{a}{b}}=n \log_m(n)=\frac{a}{b} b\log_m(n)=a log_m(n^b)=a \implies m^a=n^b$$2 Hint: (5)^{2015}(2)^{2015} = (10)^{2015}. (2)^{2018} = 2^{2015} \cdot 2^3 2 Taking logs to any base,$$x^3\log2=x^2\log3\ ,$$and therefore either x=0 or x=(\log3)/(\log2). Note that by the "change of base" formula, the last expression is the same no matter what base you are using. 2 Take natural log$$ 2^{x^3} = 3^{x^2} \implies x^3 \ln 2 = x^2 \ln 3 \implies x^2 \left( x\ln2 - \ln 3\right) = 0 \implies x_{1,2} = 0,\ x_3 = \frac {\ln 3}{\ln 2} = \log_2 3 $$2 Think modulo 10: The powers of 7 modulo 10 are, in order, 1,7,9,3, and then it repeats. Specifically, 7^7\equiv_{10} 3. Now, the powers of 3 modulo 10 are 1,3,9,7, and then it repeats. Specifically, 3^7\equiv_{10}7. We see that for each time we take the seventh power, the last digits alternates between seven and three. We start with 7 ... 2 Last digit for 7^7 = 823543 is 3 Last digit for (7^7)^7 is 7 Last digit for ((7^7)^7)^7 is 3 .... Last digit for 7^{th} power taken odd number of times is 3 Last digit for 7^{th} power taken even number of times is 7 Last digit for 7^{th} power taken 100 times should be 7 2 The problem boils down to comparing different values of the function f(x)=x^{\frac{1}{x}}, or, equivalently, different values of:$$ g(x)=\frac{\log x}{x}\qquad \text{or}\qquad h(t)=t\, e^{1-t}. \tag{1}$$g(x) a maximum at x=e and h(t) has a maximum at t=1. By computing the Taylor expansion of h(t) at t=1, we have:$$ h(t)\approx ...

2

Your calculator (and Alpha) are correct: $-1^2 = -(1^2) = -1$ whereas $(-1)^2=+1$. When evaluating expressions with multiple operations you have to follow the proper order of operations (order of precedence). Quickly: Exponentiation then Multiplication/Division then Addition/Subtraction. The reasoning behind this ordering has to do with the complexity of ...

2

You have it wrong. $$2\uparrow\uparrow\uparrow\uparrow2=2\uparrow\uparrow\uparrow2=2\uparrow\uparrow2=4$$ In fact, $2 \uparrow^n 2= 4$ for any $n$. However $2\uparrow\uparrow3=16$ and $2\uparrow\uparrow\uparrow3=65536$. $2\uparrow\uparrow\uparrow\uparrow3=2\uparrow\uparrow\uparrow4=2\uparrow\uparrow65536$. This is an stack of exponentation with height ...

Only top voted, non community-wiki answers of a minimum length are eligible