New answers tagged

0

If you do not want to use Lambert function or any numerical method, you can have explicit approximations using Newton-like methods of high order $n$ starting at $x=1$. For example $$x_{(2)}=\frac{5}{1+e}$$ $$x_{(3)}=\frac{10+14 e-e^2}{2+8 e+e^2}$$ $$x_{(4)}=\frac{60+296 e+32 e^2-4 e^3}{12+116 e+56 e^2+2 e^3}$$ $$x_{(5)}=\frac{720+8016 e+4344 e^2-120 e^3-18 e^...


1

Hint: Let $|x|=a\ge0, \sqrt{x^2-1}=b\implies a^2-b^2=1, a^2+b^2=?$ If $x\ge0, x=a$ $$(a+b)^a+(a-b)^a=2(a^2+b^2)=(a+b)^2+(a-b)^2$$ $$\implies (a+b)^a+\dfrac1{(a+b)^a}=(a+b)^2+\dfrac1{(a+b)^2}$$ If $p+\dfrac1p=q+\dfrac1q, p=q$ or $p=\dfrac1q$ What if $x<0?$


0

Hint: Notice that with exponent $2$, $$(|x| + \sqrt {x ^ 2 - 1}) ^ 2 + (|x| - \sqrt {x ^ 2 - 1}) ^ 2= 2(2x^2 - 1),$$ so $x=2$ is a solution. And as $$(|x| + \sqrt {x ^ 2 - 1}) (|x| - \sqrt {x ^ 2 - 1})=1,$$ the function is even, so $-2$ is another solution. With exponent $1$, $$2|x|=2(2x^2-1)$$ and $|x|=1$ is also a solution. Remains to show if they are the ...


1

Consider $A:=|x|+\sqrt{x^2-1}$; then we have, $$A^{-1}=\frac{1}{|x|+\sqrt{x^2-1}}=\frac{|x|-\sqrt{x^2-1}}{x^2-(x^2-1)}=|x|-\sqrt{x^2-1}$$ So, your equation is essentially, $$A^x+A^{-x}=2(2x^2-1)$$ and note that we have $A+A^{-1}=2|x|$ and $A-A^{-1}=2\sqrt{x^2-1}$ Can you take it from here?


0

I prefer to do this this analytical process. Look at the function with lower k: I mathematica it is often sensible to look at the function sequence on the borders of the interval of interest. In this case it is possible to state this values for all k. $f(0)=1$ and $f(0.5)=0$. Inbetween the function can either be bigger or smaller than $1$. The functions are ...


0

$$e^0=1\text{ vs. }0^y=0$$ precludes any solution of the form $$e^x=x^y.$$ (And as shown by @nala, no polynomial can do. We can also add that no fractional polynomial can do. Furthermore, it is know that $e^x=1+x+\frac{x^2}2+\frac{x^2}{3!}+\cdots$ with an infinity of terms.) The equation $$e^x=x^y$$ is simply solved by $$y=\frac{\ln x}x$$ but this is of no ...


1

There's no $y$ such that $e^x = x^y$ for all $x$, because in the long term, the exponential beats any polynomial. One way to see it, is when defining the exponential as a power series: $e^x := \displaystyle\sum_{n \geq 0} \dfrac{x^n}{n!}$. In particular, if the expression you're trying to simplify has $x \approx 0$, you can use the expression above and ...


0

As stated by others, this simply boils down to order of operations (you can use your favourite acronym to help you, such as PEDMAS, or in the UK: BIDMAS). So with ${-2^2}$, we have no Parenthesis, but we do have an Exponent, so we calculate this part first. ${2^2 = 4}$. Hence ${-2^2=-4}$. With ${(-2)^2}$, we do have Parenthesis, then an Exponent - so ${(-2)^...


0

You're not alone in this confusion. It's very common to see $-2^2$ and read it the way you would read the symbols left to right, i.e."negative two squared" as the order of operations. However, remembering PEMDAS (parenthesis, exponents, multiplication, division, addition, subtraction) what $-2^2$ really means is "the negative of two squared&...


1

I have found a counterexample: $$a=\frac{2}{5},\ b=\frac{3}{10},\ c=\frac{3}{10},\ n=1.$$ This gives: $$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n} = \left(\frac{2}{5}\right)^{\frac{6}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\frac{3}{10}\approx 1.0037 >1.$$


0

$17^2 =289$ $\Rightarrow$ $17^2 =-1[5]$ and $ 17^2 =1[2]$ $(17^{2}) ^8=1[5] $$\Rightarrow$$ 17^{16} =1[5]$ So $17^{16} =1[5]$ and $17^{16}=1[2] $ $gcd(5,2)=1$ $\Rightarrow$ $17^{16} =1[10]$ Note:$a=b[n] $ and $a=b[m] $ and $gcd(m, n) =1$$\Rightarrow$ $a=b[mn]$


1

Yes you can. From definition, for some $k\in\mathbb Z$: $$\frac {n_1n_2-1}2+2k=\frac{n_1-1}2+\frac{n_2-1}2$$ Hence: $$(-1)^{(n_1-1)/2+(n_2-1)/2}=(-1)^{2k+(n_1n_2-1)/2} = (-1)^{2k}(-1)^{(n_1n_2-1)/2}=(-1)^{(n_1n_2-1)/2}$$


1

Hint: Prove that $$(-1)^a \equiv (-1)^b \Leftrightarrow a\equiv b \pmod{2}.$$


1

My computer says: $$10124^{2}=102495376$$ $$10128^{2}=102576384$$ $$10136^{2}=102738496$$ $$10214^{2}=104325796$$ $$10278^{2}=105637284$$ $$11826^{2}=139854276$$ $$12363^{2}=152843769$$ $$12543^{2}=157326849$$ $$12582^{2}=158306724$$ $$12586^{2}=158407396$$ $$13147^{2}=172843609$$ $$13268^{2}=176039824$$ $$13278^{2}=176305284$$ $$13343^{2}=178035649$$ $$...


0

I have a second sketch/Partial proof (tell me if i'm wrong) : We want to show Let $0.65\leq x<1$ and $1\leq k\leq n$ two naturals numbers with $n\geq 10^{10}$ then we have :: $$P(k)=(1-x)^{(2x)^{1+\frac{k}{n}}}+x^{(2(1-x))^{1+\frac{k}{n}}}\leq 1\quad (I)$$ We use a form of the Young's inequality or weighted Am-Gm : Let $a,b>0$ and $0<v<1$ then ...


0

In the case of radioactive decay, the amount of non-decayed substance is $$f(t)=A2^{-\frac{t}{T}}$$ where $A$ is the initial quantity, $t$ is time, and $T$ is the half-time of the decay. Plugging $t=0$ yields $f(0)=A$, the initial quantity, as expected.


4

Yes, it does. If you have an exponential function to, for instance, describe a population that doubles daily and starts at $a$ $$p(x) = a2^x$$ then taking $2^0$, should naturally equal $1$, because at time $0$, the population is exactly the starting population $a$.


1

Here is general way. You need to solve this linear equation $$\dfrac {a^{12x}}{a^{7y}}=a \Longrightarrow 12x-7y=1$$ where $x,y\in \mathbb {Z^+}$ Now, $\gcd (12,7) =1$ , then we always have a solution. Thus, it is not even necessary to solve this equation with the Euclidean algorithm.


0

Using the Bezout relation $1=12\times3-7\times5$, we have $(a^{12})^3/(a^7)^5=a\in\mathbb Q$.


2

Your solution is fine. You could do it in one line ... \begin{eqnarray*} a=\frac{(a^{12})^3}{(a^7)^5}. \end{eqnarray*}


3

Hint: $\gcd(7, 12) = 1$ so done.


1

Well, we could the same things more directly: write $a$ as $\frac{(a^{12})^3}{(a^7)^5} = \frac{a^{36}}{a^{35}}$, or as $\frac{(a^7)^7}{(a^{12})^4} = \frac{a^{49}}{a^{48}}$, and conclude that it's in $\mathbb Q$. Your work probably simplifies to one of these after some substitutions.


1

Not a final answer but too long for a comment. Hoping that you do not mind, I shall define $$f(x)=x^x\log\left(\Gamma(x+1)\right)-\log\left(\Gamma(x^x+1)\right)$$ Close to $x=1$, we have $$f(x)=\left(\frac{\gamma }{2}+\frac{\pi ^2}{12}-1\right) (1-x)^3+$$ $$\frac{1}{12} (1-x)^4 \left(8-2 \gamma -\pi ^2-4 \psi ^{(2)}(2)\right)+O\left((1-x)^5\right)$$ which ...


1

You have the right idea, but I'm a bit confused by your description of the domain. Surely the union of those two sets is simply $$\Omega=\{re^{i\theta}:\ r>0,\ \tfrac13\pi<\theta<\tfrac23\pi\}.$$ Either way, your reasoning from here is correct, and then the image $f(\Omega)$ can similarly be described simply as $$f(\Omega)=\{re^{i\theta}:\ r>0,\ ...


1

If you know the fundamental theorem of arithmetic, then you should be able to show the following lemma: A positive integer $k$ is a perfect $n$-th power if and only if for every prime number $p$, the exponent of $p$ appearing in the factorization of $k$ is a multiple of $n$. Now if $k$ is a perfect $m$-th power and a perfect $n$-th power, then by the lemma,...


1

Hint: $(m,n)=1 \Rightarrow \exists x, y \in \mathbb{N}, mx-ny=1.$ Can you start from here?


0

Using only 'ground floor' modular arithmetic theory, you can build a bottom-up presentation of exponent relations of the $\text{modulo-}71$ structure to answer this question. To get things moving, you solve $2x = 1 \pmod{71}$ and find that $\large 2^{-1} = 2^2 \cdot 3^2 \pmod{71}$. So, $\quad 2^{35} \equiv 1 \pmod{71} \; \text{ iff}$ $\quad 2^{34} \equiv 2^2 ...


1

Here's another way: $$ 2^{36}\equiv 64^6 \equiv (-7)^6 \equiv (-343)^2 \equiv 12^2 \equiv 2 \pmod{71} \Rightarrow 2^{35} \equiv 1 \pmod{71} $$


1

Since $71\equiv3$ mod $4$, $k$ is a quadratic residue if and only if $71-k$ is a nonresidue. In particular, $70=71-1$ and $35=71-36$ are nonresidues. But since $70=2\cdot35$, we can conclude that $2$ is a quadratic residue, i.e., $2\equiv a^2$ mod $71$ for some $a$, in which case $2^{35}\equiv a^{70}\equiv1$ mod $71$ by Fermat's little theorem. Remark: This ...


3

According to Euler's criterion, $2^{35}\equiv\left(\dfrac2{71}\right)\bmod71$. Furthermore, $\left(\dfrac2{71}\right)=1$, because $71\equiv-1\bmod8$.


3

$$2^{35} = 2^{10} \times 2^{10} \times 2^{10} \times 2^5 \\ = 1024 \times 1024 \times 1024 \times 32 \\ \equiv 30 \times 30 \times 30 \times 32 \equiv 1 \mod 71.$$


1

The answer is no. The best you could do is as follows. We have $$ 10^{10^{100}} \cdot 10^{10^{10^{55}}} = 10^{10^{100} + 10^{10^{55}}} $$ If you like, we could "simplify" the exponent a bit by writing $$ 10^{100} + 10^{10^{55}} = 10^{100}[1 + 10^{10^{55} - 100}], $$ but this exponent is certainly not equal to $10^{10^{57}}$.


1

I am assuming $a>0,n>1$. Keep $1-(ax+1)^{1-n}=u,du=a(n-1)dx/(ax+1)^n$ which gives$$I=\int_{0}^1\frac{u^{m-1}}{a(n-1)} du=\begin{cases}\frac1{am(n-1)},&m>0\\\infty,&m\le0\end{cases}$$ For $n=1$ the integrand is $0$. For $n<1$ we get $-\infty$ for $m\le0$ but the same result as above for $m>0$.


0

Yes the natural log of $e^{x}$ is $x$ and $e^{\ln{x}}=x$ so it is reasonable to say that taking the exponential (to base $e$) is the same as the antilog (to base $e$). The inverse of log to base $10$ (for example) is to raise $10$ to the power of $x$.


2

By the laws of indices/logarithms, $$\exp H(N)= \exp\Big(-\sum_{N} w_i\log w_i\Big) = \prod_{N} \exp(-w_i\log w_i) = \prod_{N} \exp(\log(w_i^{-w_i})) = \prod_N {w_i}^{-w_i},$$ since $\exp$ and $\log$ are inverses of each other.


1

See that $2^{16^x} = 16^{2^x}$ is $2^{2^{4x}} = 2^{2^{x+2}}$ then $4x = x + 2$ and get $x = 2/3$.


2

Alternative approach $$16^{2^x} = [2^4]^{2^x} = 2^{4 \times 2^x}.$$ Since this is equal to $2^{16^x}$ you have $$16^x = 4 \times 2^x \implies 8^x = 4 \implies x = (2/3).$$


1

Take log on base 2, then $$16^x \log_2 2=2^x \log_2 16. \implies 16^x= 4. 2^x$ \implies 2^{4x}=2^{x+2} \implies 4x=x+2 \implies x=2/3.$$


0

It turns out that my Professor is making a mistake. The correct system of equations is $$ \begin{cases} 3^{x + y} + 2^{y - 1} = 239, \\ 3^{2x - 1} + 2^{y + 1} = 43. \end{cases} $$ Answer: We know that $ 239 = 243 - 4 = 3^5 - 2^2 $ and $ 43 = 27 + 16 = 3^3 + 2^4 $, so we have new system of equations that satisfy $$ \begin{cases} x + y = 5, \\ y - 1 = 2, \\ 2x ...


1

Comment: Finding by plotting the equations, using Wolfram we get following figure: $(x, y)≈ (2.2, 1), (-2, 4.4)$


4

Hint: $$(3^x)^2+(12\cdot3^y)3^x-405=0$$ The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$ For rational $3^x,$ we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say) $$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$ $$\implies d+2\cdot3^y\ge\sqrt{45}>6$$ Again, $d+...


0

The statement is equivalent to $$ x^\varepsilon > \log x, $$ for any $\varepsilon>0$ and sufficiently large $x$. From the power series expansion of the exponential function $e^x>x$ for all $x>0$, i.e., $x>\log x$ for all $x>0$. Thus, $$ \log x = \frac{2}{\varepsilon }\log (x^{\varepsilon /2}) < \frac{2}{\varepsilon }x^{\varepsilon /2} ...


1

Write $$x^k=x \log(x)\implies x^{k-1}=\log(x)\implies x^{k-1}=\frac 1{k-1}\log \left(x^{k-1}\right)$$ Let $y=x^{k-1}$ to make $$y=\frac 1{k-1}\log(y)\implies y=\frac{W(1-k)}{1-k}$$ where $W(.)$ is Lambert function. In the real domain, this function is defined as long as $1-k \geq -\frac 1e$ that is to say $k \leq 1+\frac 1e$. Back to $x$, the curves ...


0

Yes, it is a basic result in asymptotic analysis that for any $\alpha,\beta >0$, $$\ln^\alpha x=_{\infty}o\bigl(x^\beta\bigr). $$


1

Then do $$\left (x^{a/b}\right )^b=2^b,$$ gives you the equation $x^a=2^b.$ This one you know how to solve.


4

If $k > 1$, then $k = 1 + \varepsilon$ for some $\varepsilon > 0$. So $x^k = x^{1 + \varepsilon} = xx^\varepsilon$. Thus, to show $x^k = xx^\varepsilon > x\log(x)$, eventually, amounts to showing $x^\varepsilon > \log(x)$, eventually. And, this latter fact is true because: \begin{align*} \lim_{x \to \infty}\frac{\log(x)}{x^\varepsilon} &= \...


0

Induction will work great, but I'll mention that you might be getting a bit confused or distracted by the variables they chose to use. I'm curious if you find $$\dfrac{d^n}{dx^n}e^{ax}=a^n e^{ax}$$ a bit more obvious. Because it's the exact same thing as your question.


2

First find the Jordan matrix of $A$. The eigenvalues of $A$ are found from the characteristic equation $$0=\det\begin{pmatrix}-1-\lambda & x & -x \\ 1 & -1-\lambda & 0 \\ 1 & 0 & -1-\lambda \end{pmatrix}=-(1+\lambda)^3$$ to be $\lambda=-1$ thrice. There is only one associated eigenvector $(0,1,1)$. A generalized eigenvector is found ...


1

Because$$\frac{x^{22-r}}{x^r}=x^{22r-2r-r}=x^{22-3r}\ne x^{\frac{22-2r}r}.$$


2

Consider the collection, $\mathscr{C}$, of all subsets of $C$ which sum to $≥b^k$. We know that $\mathscr{C}$ is not empty (since it at least contains $C$). Let $C_0$ be a subset with minimal sum in $\mathscr C$. We claim that the sum of the coins in $C_0$ is exactly $b^k$, which will clearly conclude the proof. So, suppose otherwise. Suppose $$N=\sum_{c\...


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