5

When you square the quantity $x$, the negative sign disappears, and it doesn't come back if you raise it to the $\frac{1}{6}$ power. If you just take $\frac{1}{3}$ power, then the negative stays negative. $((-1)^2)^\frac{1}{6}=1^\frac{1}{6}=1$, while $(-1)^\frac{1}{3}=-1$ Therefore, what you see on the first graph is the absolute value of the second graph. ...


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} &= \...


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+...


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 $a$.


3

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


3

Note the limit is only defined as $x\to 0^+$, so I'll answer as such. Not sure how 'clever' this is either and the solution is a little fast and loose, but oh well. Using LHR or other methods we have $$ \lim_{x\to0^+} f_1(x)^{f_2(x)} = 1, $$for $f_1,f_2\in \{\sin,\operatorname{id}\}$. This means we can interchange some of the exponents with mild impunity. ...


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

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


3

Since $(x,y)$ are roots of $\ x^2-sx+p=0$ Then consider this equation as the characteristic equation of the linear induction relation $$U_{n+1}=sU_u-pU_{n-1}$$ Which solution is $\ U_n=x^n+y^n\ $ with initial values $\begin{cases}U_0=2\\U_1=4\end{cases}$ Then go on calculating successive values: $U_2=4U_1-2U_0=12$ $U_3=4U_2-2U_1=40$ $U_4=4U_3-2U_2=136$ $U_5=...


2

Another way is to write out $$ (x+y)^3=(x^3+y^3)+3xy(x+y), $$ $$ (x^3+y^3)^2=(x^6+y^6)+2(xy)^3. $$


2

$x^2+y^2=(x+y)^2-2xy=16-4=12.$ $x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$. Now you can use your factorization: $x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.


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 ...


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\...


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 $$16^x = 4 \times 2^x \implies 8^x = 4 \implies x = (2/3).$$


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.


2

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


1

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


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?


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 ...


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 $m>0$.


1

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


1

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


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.


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 ...


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