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9

The notion of multivalued exponentiation makes the most sense once you learn complex exponentiation. In complex numbers, $x^y$ with $x\neq 0$ has $q$ values if $y=p/q$ is a rational number in reduced form, and infinitely many value otherwise. So $2^{2}$ has one complex value, $2^{3/7}$ has $7$ complex values, $2^{\sqrt{2}}$ and $2^{1+i}$ both have ...


8

Theorem of Siegel that for real $\lambda$ and distinct primes $p,q,r,$ the numbers $$ p^\lambda, \; q^\lambda, \; r^\lambda, $$ cannot all be rational unless $\lambda$ is an integer. See page 455 in Alaoglu and Erdos, On Highly Composite and Similar Numbers (1944), also chapter 2 of Introduction to Transcendental Numbers by S. Lang. I see there is ...


4

Not an answer. Just a related plot I produced and feel like sharing. This seems to very much support the model that in $2^n$ there are $n\log_{10}2$ random digits. In the figure below the blue cloud of dots gives the difference $S(n)-P(n)$, where $S(n)$ is the actual sum of base ten digits of $2^n$, and $P(n)=\frac92n\log_{10}2$ is the predicted sum of ...


3

Estimation Looking at $$ 2^n = 10^x \iff x = n \log_{10} 2 = \frac{\ln 2}{\ln 10} n \approx 0.3 \, n $$ $(2^n)_2$ has $n+1$ digits, $(2^n)_{10}$ has about $n/3$ digits. $10$ new digits in base $2$ ($2^{10} = 1024$) give about $3$ new digits in base 10. My gut feeling is that with more and more digits, more and more digits have the chance to be non-zero, ...


3

I would say that there is no such $\alpha$. In fact I would bet my house on it. It would be astonishing if (say) $S(2^n)$ were found to be eventually monotonic increasing. But I don't think anybody has a proof that it's not. In the other direction: Does $S(2^n)$ even tend to infinity as $n \to \infty$? Again, it would be astonishing if it didn't. But I ...


3

By the little Fermat theorem we have $$5^{10}\equiv 1\pmod{11}$$ and since $$9^{{13}^{17}}\equiv-1\pmod{10}$$ then we see that $$5^{9^{{13}^{17}}}\equiv 5^{-1}=9\pmod{11}$$


3

I am assuming you are only looking for real solutions. Even if not, this answer can still be useful. We can see that for there to be a real solution, $a$ must be in the form $x^\frac{1}{x}$. Using calculus to maximize this gives that the maximum value of $x^\frac{1}{x}$ and thus the maximum value of $a$ is $e^\frac{1}{e}$. From this you get that the real ...


3

\begin{align} e^{-4\ln\left(x\right)+8\ln\left(y\right)+2}&=e^{-4\ln x}e^{8\ln y}e^2\\\\ &=e^2\frac{\left(e^{\ln y}\right)^8}{\left(e^{\ln x}\right)^4}\\\\ &=\displaystyle\boxed{\displaystyle\frac{e^2y^8}{x^4}} \end{align}


2

Take log. $$x\log5+y\log3=\log45$$ $$x\log3+y\log5=\log75$$


2

$$\left(\frac{1}{\sqrt{x}}\right)^{11} = \left(x^{-1/2}\right)^{11} = x^{-11/2} \neq x^\sqrt{11} = x^{11^{1/2}}$$ So for equality you would need $\frac{-11}{2} = \sqrt{11}$ which is not possible.


2

Those are not the same. $\left(\frac1{\sqrt x}\right)^{11}=(x^{-1/2})^{11}=x^{-11/2}$.


2

Every term has a factor of $5^x$ which is factored out. Also, note that: $$5^{-1} = \frac 1{5^1} = \frac 15,\quad 5^2 = 25$$ $$\begin{align} 5^{-1}\cdot \color{blue}{5^x} - (1)\cdot \color{blue}{5^x} - 5\cdot \color{blue}{5^x} + 5^2\cdot \color{blue}{5^x} &= \frac 15\cdot \color{blue}{5^x} -(1)\cdot \color{blue}{5^x} - 5\cdot \color{blue}{5^x}+ ...


2

Using the Chinese Remainder Theorem in addition to other bits. Observe that your modulus factors like $m=2^8\cdot3\cdot 5^9$. Your number is very obviously divisible by $2^8$, so we can forget about that factor until the end. Modulo $3$? The number $2^{2^{\text{ZILLION}}}$ is clearly a power of $4$, so its remainder modulo $3$ is equal to $1$. Modulo ...


2

There are no algebraic "closed-form" expressions for the solutions. However, we can rewrite it as $$\begin{align*} -a^x&=-x\\ -1&=(-x)a^{(-x)}\\ -1&=(-x)e^{\ln(a)\cdot (-x)}\\ -\ln(a)&=(-\ln(a)x)e^{(-\ln(a)x)}\\ -\ln(a)&=ye^y,\quad\text{where } y=-\ln(a)x. \end{align*}$$ The solution of this equation is, by the definition of the Lambert W ...


2

See MO It follows from the Six Exponentials Theorem. Apparently.


2

You can forget about $a$ and $b$ and concentrate on proving that $c^{4d+1}-c$ is a multiple of $10$. Since $10=2\cdot 5$, you only need to prove that $c^{4d+1}-c$ is a multiple of both $2$ and $5$. It is easy to see that $c^{4d+1}-c$ is a multiple of $2$ since even $\cdot$ even is even and odd $\cdot$ odd is odd. It is easy to see that $c^{4d+1}-c$ is a ...


2

If the question is "Is this correct?", the answer is yes. :-)


2

Suppose that you have $a^2$, then: $$4 \cdot a^2 = 2^2 \cdot a^2= (2 \cdot a)^2=(2a)^2$$


2

There is a printing error; usually one write $\sqrt{8} = 2\sqrt{2}$, so $\sqrt{8r^2} = 2\sqrt 2 r $ may easily have become $2\sqrt{2r}$


2

$2^{1} = 2$ $2^{2} = 4$ $2^{3} = 8$ $2^{4} = 16 \rightarrow 7$ $2^{5} = 32 \rightarrow 5$ $2^{6} = 64 \rightarrow 10 \rightarrow 1$ Then.. $2^{7} = 128 \rightarrow 11 \rightarrow 2$ $2^{8} = 256 \rightarrow 13 \rightarrow 4$ $2^{9} = 512 \rightarrow 8$ $2^{10} = 1024 \rightarrow 7$ $2^{11} = 2048 \rightarrow 14 \rightarrow 5$ $2^{12} = 4096 ...


2

Nobody in the world knows how to do this faster than calculating the decimal representation of $2^n$ and adding up all its digits. When $n$ gets big, the challenge is to calculate $2^n$ as fast as possible; Fast Fourier Trasforms are the way to go, but this is a deep subject. Start with Wikipedia's article on Karatsuba's algorithm for a gentle introduction ...


1

$2^n$ has $d$ digits if and only if $10^{d-1} \leq 2^n < 10^d$. Taking $\ln$ gives $(d-1) \ln 10 \leq n\ln 2 <d \ln 10$ which gives $d-1 \leq n \frac{\ln 2}{\ln 10} < d$, showing that $d-1$ is equal to the integer part (so-called floor) of $n\frac{\ln 2}{\ln 10}$. The number of digits of $2^n$ is therefore equal to one plus the integer part of ...


1

First notice that $a^{\frac{1}{b}} = \Bigg(1 + \frac{b}{x}\Bigg)^{\frac{x}{b}} $. Now $$\begin{align}\ln a = \ln\Bigg(1 +\frac{b}{x}\Bigg)^x &\Rightarrow \Bigg (1 + \frac{b}{x}\Bigg)^{\Big(1 + \frac{x}{b}\Big)}\ln a = \Bigg (1 + \frac{b}{x}\Bigg)^{\Big(1 + \frac{x}{b}\Big)} \ln\Bigg(1 +\frac{b}{x}\Bigg)^x\\&\Rightarrow \Bigg[\Bigg (1 + ...


1

Because $4 = 2\cdot 2$ and multiplication works like this: $$ 6^2 \cdot 4 = 6\cdot 6 \cdot 4 = 6\cdot 6 \cdot 2\cdot 2 = 6\cdot 2\cdot 6\cdot 2 = (6\cdot 2)^2$$


1

Hint: since $$F_{n}=\dfrac{1}{\sqrt{5}}\left(\alpha^n-\beta^n\right),\alpha=\dfrac{1+\sqrt{5}}{2},\beta=\dfrac{1-\sqrt{5}}{2}$$ so $$\dfrac{F_{kn}}{(F_{n})^k}=\dfrac{\frac{1}{\sqrt{5}}(\alpha^{nk}-\beta^{nk})}{\left(\dfrac{1}{\sqrt{5}}\left(\alpha^n-\beta^n\right)\right)^k}=\sqrt{5}^{k-1}\dfrac{\alpha^{nk}-\beta^{nk}}{(\alpha^{n}-\beta^{n})^k}\to ...


1

$$\lim_{n\to\infty}\dfrac{F{kn}}{{F_{n}}^k}=\dfrac{\dfrac{\left({\dfrac{1+\sqrt5}{2}}\right)^{nk}-\left({\dfrac{1-\sqrt5}{2}}\right)^{nk}}{\sqrt5}}{\left({\dfrac{\left({\dfrac{1+\sqrt5}{2}}\right)^{n}-\left({\dfrac{1-\sqrt5}{2}}\right)^{n}}{\sqrt5}}\right)^k}=\sqrt5^{-1+k}\cdot1=5^{\dfrac{k-1}{2}}$$


1

Since natural number implies $a,b,c,d\ge1$ $$(10^ab+c)^{4d+1}=\sum_{k=0}^{4d+1}(10^ab)^kc^{4d+1-k}=c^{4d+1}+\underbrace{\sum_{k=1}^{4d+1}(10^ab)^kc^{4d+1-k}}_{\text{divisible by }10}$$ And since cyclicity of all natural numbers is a factor of 4,i.e. of $\{1,2,3,4,5,6,7,8,9\}$ are $\{1,4,4,2,1,1,4,4,2\}$; $c^{4d+1}$ ends in c and after subtracting c the ...


1

We don't need an analytical approach to disapprove this equality. If we can find a counter example for $x\in\mathbb{R}^+$ where the left doesn't equal the right, we will be done. Take $x$ to be $4$. Then $\frac{1}{2^{11}}\ll 1$. The right hand side $4^{\sqrt{11}}\gg1$. So the equality doesn't hold. Another approach would be to take a significantly large $x$. ...



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