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Every non-zero complex number $z$ has exactly two complex square roots - this is a consequence of the field of complex numbers being algebraically closed (Wikipedia link). If $$z=re^{i\theta}=r\cos(\theta)+ri\sin(\theta)$$ then the square roots of $z$ are \begin{align*} \sqrt{r}e^{i\theta/2}&=\sqrt{r}\cos(\theta/2)+\sqrt{r}\,i\sin(\theta/2)\\ ... 11 Maybe you want to solve z^2=i. It is easy to verify that z=\pm\frac{\sqrt2}{2}(1+i) satisfy the equation. Generally, every polynomial with complex coefficient has a root in \mathbb C, or, equivalently, complex field has no algebraic field extension. 11 The following (not particularly elegant) proof uses reasonably basic multiplication and division. We need to show that 7^{31} > 8^{29}, i.e. that \dfrac{7^{31}}{8^{29}}>1. We have: ... 9 HINT: Divide by 4^x to geta^2-a-2=0$$where a=\left(\dfrac32\right)^x Can you solve for a? Now for real x,a>0 See also : Exponent Combination Laws 8 The square root function is not as nicely behaved on the complex numbers, \Bbb C as it is on the (nonnegative) real numbers, [0, \infty). It's true that we can find exactly two solutions to the equation w^2 = z for any nonzero z \in \Bbb C, but unlike in the usual real setting, we cannot make choice of w that continuously depends on z, or more ... 7 Take logs, to get b\ln a>a\ln b, or \frac{\ln a}a>\frac{\ln b}b. The function \frac{\ln x}x increases to a maximum at x=e=2.71828..., then decreases. So your question is true if e<a<b If you don't like logs, take the ab^{th} root of both sides, and get \sqrt[a]a>\sqrt[b]b when e<a<b. 7 so start with for a>1 and \forall n big enough we have$$n^a < a^n \Leftrightarrow \log(n^a)<\log(a^n)\Leftrightarrow a\log(n)<n\log(a)\Leftrightarrow 1<\frac{\log(a)}{a}\frac{n}{\log(n)} $$and we fixed a, so \frac{\log(a)}{a}=c>0 is just a constant. So in fact we have to show, that$$ 1<c\frac{n}{\log(n)} $$holds for all ... 7 One should not say "equals undefined"; one should say "is undefined". The is the "is" of predication, not the "is" of equality. 0^0 is indeterminate in the sense that if f(x) and g(x) both approach 0 as x\to a then f(x)^{g(x)} could approach any positive number or 0 or \infty depending on which functions f and g are. But in some ... 7 Let r be the random number, and let r_n be the number obtained after repeating the 4 step procedure n times. Let b be the base used in steps 1 and 3. Starting with r_n, we do the following: 1 Exponentiate: b^{r_n} 2 Raise to the power x: (b^{r_n})^x = b^{xr_n} 3 Take the log (same base): \log_b(b^{xr_n}) = xr_n 4 Take the x-th root: ... 6 Others may bristle at this "proof," but:$$7^{31} = 157,775,382,034,845,806,615,042,743 \\ 8^{29} = 154,742,504,910,672,534,362,390,528$$If all else fails, just calculating the expressions and comparing them will work. This particular problem is only mildly tedious to attack this way if you have pen/paper. 6 You can rewrite this as$$(e^x)^2-6-e^x=0(e^x+2)(e^x-3)=0$$I'll let you take it from here 6 HINT : Multiply the both sides by e^x to get$$(e^x)^2-6-e^x=0.$$Now, let e^x=t. 5 let e^x = a$$ a - \frac{6}{a} - 1 = 0 a^2 - a-6 = 0 a = 3 \ or \ -2  e^x = 3$$Edit: for x \in \mathbb{R} Can you find the value of x now? Hint: take \ln of both sides. 5 It's the same trick, except easier. Pull out all the 2's and 5's to see how many 10's you get:$$ 15^{80}28^{60}55^{70} = 5^{80}4^{60}5^{70}(3^{80}7^{60}11^{70}) = 2^{120}5^{150}(3^{80}7^{60}11^{70}) = 10^{120}(5^{30}3^{80}7^{60}11^{70}) $$5 To clarify one thing (Michael already posted a great answer). You don't prove something is undefined. Undefined is not some mystical object. Undefined means we have not defined it. You can prove something can't be defined in a way consistent with other math. I can define 0^0 anyway I like. I define 0^0=54. There, now it is defined, but it is not ... 4 HINT: multiply both sides by e^x to turn it into a quadratic equation in e^x. 4 Basically instead of degrees and radians, on a complex plane we can use multiplication to express rotations. Multiplying by 1 takes you take to the same place, so it's the equivalent of 360 degrees. Multiplying by -1 is the same as 180 degrees. Multiplying by i is the same as a 90 degree rotation and multiplying by \sqrt{i} is the same as rotating ... 4 The more elegant way to solve this equation was given by lab bhattacharjee, but I'll simply elaborate on how to solve it using your method, which was still correct. Hint: We have that$$a^2 - ab -2b^2 = 0 \iff b = \frac{a}{2}, \text{or }b= -a.$$You now have a system of two equations, \log_3 a = \log_2 b and from this you can see that neither a or ... 4 This is what happens when you apply finite differences to any sequence. Here is some useful notation. If a_n is a sequence, its forward difference is the sequence$$\Delta a_n = a_{n+1} - a_n.$$(The notation should not be read as "\Delta of a_n," but as "the n^{th} term of the sequence \Delta a.") For example, if a_n = n^2, then$$\Delta a_n ...

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Nice observation -- you have found one of the fundamental properties of finite differences. Let me rewrite your question in a more "functional" way: Fix an abelian group $A$, written additively. (For instance, $A$ can be $\mathbb{Z}$ or $\mathbb{Q}$ or $\mathbb{C}$. Either choice works for your question, so if you are not on friendly terms with groups, you ...

3

Try to make a variable change of type: $$g(n) = f(n) + n^{\alpha} a$$. Try $\alpha$ from $0, 1, 2 ..$. Here is the case where $\alpha = 0, 1$ does not work. But $\alpha = 2, a = -\frac{c}{2}$ works! Then $g(n)$ satisfies: $$g(n) = 2g(n-1) - g(n-2)$$

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$$64/100 = .64$$ So then you have $.8^d = .64$ What does $.8 \times .8$ equal?

3

Let $f(x) = e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}-1 =\sum_{i=1}^N e^{k_i/x}-1$, and let $K = \sum_{i=1}^N k_i$. The restrictions that $x > 0$ and $k_i < 0$ are important in what follows. $f'(x) =\sum -\frac{k_i}{x^2}e^{k_i/x} =-\frac1{x^2}\sum k_ie^{k_i/x} =\frac1{x^2}\sum |k_i|e^{k_i/x}$, so $f'(x) > 0$. This means that your function has at ...

3

Consider $$A(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0 & \cdots & 0\\ \sin(\theta) & \cos(\theta) & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1\end{bmatrix}$$ Then $A(\theta)^n = A(n\theta)$. ...

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Other answers have explained why $\sqrt{i}$ is a complex number, and shown how to compute it using complex exponentials, but you can also compute it directly. If you want to find the complex number $a + bi$, where $a$ and $b$ are real numbers, such that $\sqrt{i} = a + bi$, then square both sides and solve $$i = (a + bi)^2.$$ Expanding the right-hand side, ...

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We can show that the only $nice$ solutions to $$f(x^y)=f(x)^{f(y)}\qquad(1)$$ are the identity function, and the constant functions $1$ and $-1$. There are also a few $ugly$ solutions. To avoid problems with $0^0$, let's first find all functions $f$ such that $f:(0,+\infty)\to(-\infty,0)\cup(0,+\infty)$. (Note that because the question is about nonnegative ...

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Your last step doesn't work, instead do: $$3^{3x - (4x - 2)} = 3^{x+4}.$$

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I would follow the pseudocode given here: Write your exponent, 1357, in binary: $10101001101_2$. Let $b := x\mod 2623$. Let $r := 1$. Step through the bits from right-to left: If the bit is $1$: Let $r := b \cdot r \mod 2623$. Let $b := b^2 \mod 2623$. Then $r$ will be your final result. This requires 17 multiplications modulo $2623$. In general it ...

3

It is no longer true that $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ when $a,b$ are not positive real numbers. Additional remark: One must be careful when talking of $\sqrt{\cdot}$ in the realm of complex numbers, because it cannot be defined globally, i.e. there is no continuous function $$f:\Bbb C\to \Bbb C.$$ such that $f(z)^2=z$ for all $z\in\mathbb{C}$. Formally, ...

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If we know (with suitable restrictions on $k$ and $n$) that $k^2 = n$, then we can say $$k = \sqrt{n} = n^{1/2}$$ So simply cubing both sides yields the desired $$\bbox[10px, border: solid blue 1px]{k^3 = (\sqrt{n})^3 = \sqrt{n^3}= n^{3/2}}$$ So your original answer was correct. $k$ is indeed equal to $n^{1.5}$, your error is probably found elsewhere.

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