# Tag Info

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

Hint: $$17^{30}>16^{30}=2^{4\cdot30}=2^{120}$$

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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