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8

I assume you are looking for solutions where $x,y,a,b$ are positive, none equal to $1$. Furthermore, that $a\neq b$. [If $a=b$, then any pair $y=x$ is a solution.] The equations can be rearranged to $$\left\{\begin{aligned} y\ln(x)&=x\ln(y)\\ y&=x\frac{\ln(a)}{\ln(b)}=cx \end{aligned}\right.$$ where $c=\frac{\ln(a)}{\ln(b)}\neq1$. Substituting $y$ ...


8

Hint: $9 =3^2\\25 = 5^2$ Full solution:


7

$$\log_{10}(e^{-10000}) = -10000 * (0.4342944819) = -4342.944819$$ Thus, $$e^{-1000}=10^{ -4342.944819} = 10^{-4343} 10^{ 0.05518} = 1.13548386531 \times 10^{-4343}$$


7

$$ \frac{11^{11}}{9^{11}} =\left(\frac{2}{9}+1\right)^{11} = \sum_{k=0}^{11} \binom{11}{k} \left(\frac{2}{9}\right)^k > \sum_{k=0}^5 \binom{11}{k} \left(\frac{2}{9}\right)^k = \frac{177665}{19683} > 9 $$


6

NOTE: $$(x^k)^2 \not = x^{k^2}, \text{ instead} \quad (x^k)^2 = x^{2k}$$ For the second update note that the summation/ product of exponents is valid only for real exponent. In other words $(e^{1+2ki\pi})^{1+2ki\pi} \not = e^{(1+2ki\pi)^2}$


5

$$7^2\equiv-1\pmod{50}$$ $$\implies7^{777}\equiv(7^{2})^{388}\cdot7\equiv(-1)^{388}7\equiv7$$ and $$3^{333}=3(10-1)^{166}$$ Now $$(10-1)^{166}=(1-10)^{166}\equiv1-\binom{166}110\pmod{100}\equiv1-60\equiv41$$


5

$2^{120}=(2^{10})^{12}=1024^{12}>729^{12}=(3^6)^{12}=3^{72} \tag{1}$ In general, first take the divisors of lcm of powers and compare or first compare and then take the divisors of lcm of powers, whichever makes comparison easier. I have use the technique mentioned in bold, while Roman83 has used the latter. From $(1)$ and Roman83's answer, we get $...


5

Great question, it was fun to think of an explanation :D! The problem with your calculations is that you are using multivaluedness in the wrong way. Your mistake is in the last lines $e^{1+i2k\pi}=\left (e^{1+i2k\pi}\right )^{(1+i2k\pi)}$ $e=e^{(1+i2k\pi)^2}=e^{1+i4k\pi}e^{-4k^2\pi^2}$ But $e^{i4k\pi}=1$ from Euler's identity and $e^{1+i4k\pi}=...


4

I think that MrYourMath has isolated one way to see the main issue, which is about multivaluedness of complex functions, but I would like to see if I can bring this point into more explicit contact with your reasoning. I think the problem is ultimately one about the scope of the definitions and properties you are using. In particular, you are assuming that $...


3

Expanding my comment $$T\exp(M)=T\left(\sum_{k\ge 0}\frac{M^k}{k!}\right)=\sum_{k\ge 0}\frac{TM^k}{k!}=\sum_{k\ge 0}\frac{N^kT}{k!}=\left(\sum_{k\ge 0}\frac{N^k}{k!}\right)T=\exp(N)T$$


3

Your title asks one question ("fifth last digit") and the body of your question asks a different one ("number modulo 10,000"), which is a bit confusing. Fortunately the same answer applies to both. The last five digits of the 5th, 6th, 7th,... 13th powers of 5 are 03125, 15625,78125, 90625, 53125, 65625, 28125, 40625, 03125, at which point the sequence ...


3

$3^{-4}=({3^4})^{-1}$ But $3^4=(3^2)^2=9^2=81$ So we get the answer is $81^{-1}=\frac{1}{81}$


3

Given a finite collection $\{a_1,a_2,\cdots,a_N\}$ of positive numbers, it is true that $$\log \prod_{n=1}^N a_n=\sum_{n=1}^N \log a_n$$ I.e., $$\log(a_1a_2\cdots a_n)=\log a_1+\log a_2 +\cdots+\log a_N$$ In fact, $$\log(a_1^{p_1}a_2^{p_2} \cdots a_n^{p_n})= p_1\log a_1+p_2\log a_2 +\cdots+p_N\log a_N$$


2

Well we know $3^4$ equals 81, because $3 \times 3 \times 3 \times 3 = 81$ and there is a function where any value to the power of a minus become the inverse of that number. An inverse is basically that number turned upside down. For example: $x^{-1}$ become $1/x$ and $2^{-1} $ becomes 1/2 Therefore with these two things in mind we have $81^{-1}$ Giving ...


2

Here are two examples of the square and multiply method for $5^{69} \bmod 101$: $$ \begin{matrix} 5^{69} &\equiv& 5 &\cdot &(5^{34})^2 &\equiv & 37 \\ 5^{34} &\equiv& &&(5^{17})^2 &\equiv& 88 &(\equiv -13) \\ 5^{17} &\equiv& 5 &\cdot &(5^8)^2 &\equiv& 54 \\ 5^{8} &\equiv& &...


2

For positive values of $q,p$, we have $$p^q>q^p$$ if and only if $$\ln p^q>\ln q^p$$ if and only if $$q \ln p > p \ln q$$ if and only if $$\frac{\ln p}{p}>\frac{\ln q}{q}$$ Hence we have a function we can look at to answer such questions, namely $f(x)=\frac{\ln x}{x}$. For particular cases of positive $p,q$, we simply evaluate $f(p)$ and $f(q)$. ...


2

For large $x$, no question of what to do with $0^0$ arises and we may write $x^x = \mathrm{e}^{x \ln x}$ and $a^x = \mathrm{e}^{x \ln a}$. Note that $x \ln x$ always out-grows $x \ln a$, so there is no $a$ such that $x^x \in O(a^x)$. (In fact, your observation about the infinite limit is sufficient to establish this.) However, $x^x = \mathrm{e}^{x \ln x} \...


2

Answer to an earlier version of the question which asked to construct $x^y$ instead of merely defining it: No, at least not if "any finite method" means compass and straightedge -- for example this is famously impossible when $x=2$ and $y=1/3$.


2

$\frac{3^{n-2}}{9^{1-n}}=9$ $\implies\frac{3^{n-2}}{3^{2(1-n)}}=9$ $\implies\frac{3^{n-2}}{3^{2-2n}}=9$ $\implies3^{(n-2)-(2-2n)}=3^{2} [a^{(m-n)}=\frac{a^m}{b^m}]$ $\implies3^{3n-4}=3^2$ $\implies{3n-4}=2$ $\implies n=2$ $\frac{5^{3n-3}}{25^{n-3}}$ $=\frac{5^{6-2}}{25^{2-3}} [put(n=2)]$ $=\frac{5^4}{5^{-1}}$ $=5^{4-1}$ $=5^3=125$


2

First let's combine the exponents by using the rule $(x^a)^b = x^{a\cdot b}.$ Thus we have $$(100^3)^5 = 100^{3 \cdot 5} = 100^{15}.$$ Next, we note that $100 = 10^2$. So we replace $100$ in the above equation with $10^2$ and apply the same rule.$$100^{15} = (10^2)^{15} = 10^{2\cdot 15} = 10^{30}.$$ Thus, we've expressed it with base 10. Let me know if you ...


2

You can seperate and do it $\pmod{2}$ and $\pmod{25}$ and use chinese remainder: They are both odd so their sum is even and thus $\equiv 0 \pmod{2}$. The euler function of 25 gives 20 and thus $3^{333}\equiv 3^{13}\pmod{25}$ and $7^{777}\equiv 7^{17} \pmod{25}$. Now, $7^2=49\equiv -1 \pmod{25}$. Thus $7^{17}=7^{16}\cdot 7 \equiv 7\pmod{25}$. $3^3=27\equiv ...


2

Recall that the first equation is equal to: $\dfrac{3^{n-2}}{9^{1-n}}=9 \implies \dfrac{3^{n-2}}{3^{2(1-n)}}=3^2$ This is equal to: $3^{n-2 -2(1-n)} = 3^2$


2

I always start by writing a couple of powers, in this case modulo $50$: $$\begin{align}3^1\equiv 3&\mod 50\\ 3^2\equiv 9&\mod 50\\ 3^3\equiv 27&\mod 50\\ 3^4=81\equiv 31&\mod 50\\ 3^5\equiv 93\equiv 43&\mod 50\\ 3^6\equiv 129\equiv 29&\mod 50\\ 3^7\equiv 87\equiv 37&\mod 50\\ 3^8\equiv 111\equiv 11&\mod 50\\ 3^9\equiv 33&\...


2

The reason that it isn't true is that, regrettably, the notation is not consistent. For this reason, many people avoid using $\tan^{-1}$ and use $\arctan$ instead, and so on for the other trigonometric functions. That said, $\tan^{-1}$ is logical notation, and such notation as $\tan^2$ is illogical. However, the weight of tradition and the simple convenience ...


2

The given inequality is $x!-y!-x^n \geq 0$ Observe that $x! \geq 2(x-1)!$ So if we can prove, $[(x-1)!-y!]+[(x-1)!-x^n] \geq 0$, we are done. $(1)$ $(x-1)!-y! \geq (2y-1)! -y! \geq 0$ (as $x \geq 2y$) $(2)$ $(x-1)! -x^n \geq (x-1)! -x^{(x-2)/2}$ [as $x \geq 2y, y >n \Rightarrow x \geq 2n+2 \Rightarrow n \leq \frac{x-2}{2}$] So we are left to prove $...


2

$log(abc)=log((ab)c)=log(ab)+logc=loga+logb+logc$


2

The goal is to prove that $(1 + 2/9)^{11} > 9$. As the left-hand side is approximately $9.091843$ this will be a bit tricky. The big idea in this solution is to try to exploit the fact that $(11/9)^2 = 121/81$ is just under $3/2$, since $3/2$ will be simple to work with. Start with the inequality $3^2 \times 29 > 2^8$, i. e. $261 > 256$. ...


2

This is a variant on Servaes's answer. Note that $3^5=243=2\cdot11^2+1$. Using a binomial expansion and some extremely crude upper bounds, we find $$\begin{align} 3\cdot9^{12} &=3^{25}\\ &=(2\cdot11^2+1)^5\\ &=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\ &\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+11^8+11^8\\ &\lt32\...


2

We know that $e^{i\pi} = -1$. Transforming: $a^{i\pi/\ln a} = -1$ Then: $a^{(i\pi/\ln a)+\log_a b} = -b$



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