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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)\\ ... 13 The only real solution is given by x=0, since \sqrt{2-\sqrt{3}} and \sqrt{2+\sqrt{3}} are positive real numbers with product one, hence your equation is equivalent to: e^{cx}+e^{-cx}=2 $$or:$$ \cosh(cx) = 1 $$for c=\log\sqrt{2+\sqrt{3}}. 11 Maybe you want to solve z^2=i. It is easy to verify that z=\pm\frac{\sqrt2}{2}(1+i) satisfy the equation. Generally, every polynomial with complex coefficient has a root in \mathbb C, or, equivalently, complex field has no algebraic field extension. 10 15<20, 15^{1/20}<20^{1/20} as \dfrac1{20}>0 Now, 20^{1/20}<20^{1/15} as \dfrac1{20}<\dfrac1{15} Alternatively,$$15^{1/20}<=>20^{1/15}\iff15^{15}<=>20^{20}$$Now 15^{15}<20^{15}<20^{20} 10 HINT :$$(2-\sqrt 3)^{\frac x2}+(2+\sqrt 3)^{\frac x2}-2=0\iff \left((2-\sqrt 3)^{\frac x4}-(2+\sqrt 3)^{\frac x4}\right)^2=0$$9 2^{2^{2^{2}}} = 2^{2^{4}} = 2^{16} 9 HINT:$$\left(\dfrac{3^x}{3^5}\right)^2-2\left(\dfrac{3^x}{3^5}\right)+1=0\iff\left(\dfrac{3^x}{3^5}-1\right)^2=0$$Now 3^x=3^5\iff3^{x-5}=1=3^0 As 3\ne\pm1, x-5=0 9 The idea is to realize that 77=7\times 10+7. Why is this important in the present question? It is indeed a question about the binomial theorem, it says that :$$(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^kb^{n-k} $$Now applying this with a:=7\times 10, b:=7 and n=17 we have that : ... 8 The square root function is not as nicely behaved on the complex numbers, \Bbb C as it is on the (nonnegative) real numbers, [0, \infty). It's true that we can find exactly two solutions to the equation w^2 = z for any nonzero z \in \Bbb C, but unlike in the usual real setting, we cannot make choice of w that continuously depends on z, or more ... 8$$5^{44}<5^{45}=(5^3)^{15}=125^{15}<128^{15}=(256/2)^{15}=4^{60}/2^{15}<4^{53}$$(because 2^{15}>2^{14}=4^{7}) 7 HINT :$$A(B+C)=AB+AC$$and$$2^a\times 2^b=2^{a\color{red}{+}b}.$$6 Your number is not finite. Let x_0=\sqrt{2} and x_{n+1}=x_n^{\sqrt{2}} for all n\in\Bbb Z_{\ge 0}. Then x_n\ge \sqrt{2}. Proof: x_0=\sqrt{2} and x_k\ge \sqrt{2} means x_k^{\sqrt{2}}>x_k\ge \sqrt{2}. Assume for contradiction your number is finite, i.e. \displaystyle{\lim_{n\to\infty} x_n}:=L is finite. ... 6 Hint: Substitute y=3^x to get$$y^2-2\cdot 3^5\cdot y+3^{10}=0$$6 USING COMPLEX ANALYSIS We have$$z^5=2^5\implies z=2e^{i2n\pi/5}$$for n=0, \pm 1, \pm 2.$$\bbox[5px,border:2px solid #C0A000]{\text{Thus the five roots are} \,\, 2,\, 2e^{\pm i2\pi/5},\,2e^{\pm i4\pi/5}}\tag 1$$FACTORORING A POLYNOMIAL We have x^5=32\implies x^5-2^5=0. We can then factor x-2 from the left-hand side and write ... 5 The other answers are great and comprehensive, but from the phrasing of your question I suspect they might overwhelm you at this stage. This is hardly a real answer, but I'm guessing it's more along the lines of what you're looking for. You have to find the 5^\text{th} root of 32. Just like if you have x^2 = y you take the square root of y to get ... 5 I suggest computing the value by iterated squaring.$$ 77^2 \equiv (-23)^2 = 529 \equiv 29 \mod 100 \\ 77^4 = (77^2)^2 \equiv 29^2 = 841 \equiv 41 \mod 100 \\ 77^8 = (77^4)^2 \equiv 41^2 = 1681 \equiv 81 \mod 100 \\ 77^{16} = (77^8)^2 \equiv 81^2 \equiv (-19)^2 = 361 \equiv 61 \mod 100 $$Now$$ 77^{17} = 77^{16} \cdot 77 \equiv 61 \cdot 77 \equiv ...

5

Use the substitution $u = 3^x$ to get $$u^2-2\cdot 3^5\cdot u+3^{10}= \left(u-3^5\right)^2 = 0$$ which is quadratic in $u$ and has a single real solution $u = 3^5$. So back-substituting yields $$3^x = 3^5 \iff x=5.$$

5

Well raise both numbers to the power of $20$ That is $$\large{(15^\frac{1}{20})^{20} = 15^\frac{20}{20} = 15}$$ Now $$\large{(20^\frac{1}{15})^{20} = 20^\frac{20}{15} = 20^\frac{4}{3} = 20^{1.333..}}$$ which is greater ? $\large{15}$ or $\large{20^{1.333...}}$ Clearly , it is $\large{20^{1.333..}}$ because $\large{20^{1.333} > 20^1 > 15}$ and so ...

4

That number will not be a problem. Using elementary methods, we estimate that it will have around 101 billion digits. It's easy to store 2 digits in a byte, so we can guess that the final result will occupy about 50 GB. Most computers don't have so much memory, but 50 GB is not an unusually large disk file and the intermediate results can be stored on the ...

4

Using a GCD-like approach, start by dividing out the "smaller" term: $${5^{44}\over 4^{44}}=\left(\frac54\right)^{44}\\ {4^{53}\over 4^{44}}=4^9$$ Now the new "smaller" term is $\frac54$: $${\left(\frac54\right)^{44}\over \left(\frac54\right)^{18}}=\left(\frac54\right)^{26}\\ {4^9\over \left(\frac54\right)^{18}}=\left(\frac85\right)^{18}\\ ... 4 Basically instead of degrees and radians, on a complex plane we can use multiplication to express rotations. Multiplying by 1 takes you take to the same place, so it's the equivalent of 360 degrees. Multiplying by -1 is the same as 180 degrees. Multiplying by i is the same as a 90 degree rotation and multiplying by \sqrt{i} is the same as rotating ... 4 The function \frac1x is not continuous at 0, and not only that it is not continuous, it has different limits from each side. Consider the function: f(x)=\begin{cases} -1 & x<0\\0 & x=0\\ 1 & x> 0\end{cases} It is clear that:$$\lim_{x\to0^-}f(x)=-1\neq 0=f(0)\neq\lim_{x\to0^+}f(x)=1$$There's no reason to expect that a function ... 4 Your way of asking is not so nice(because you aren't showing any work), but I'm writing the solution in any way: Write 2^x=a, 3^x=b. Then, your equation becomes 4a^2-ab-18b^2=0, which is equivalent to (4a-9b)(a+2b)=0. So, we must have either 4a=9b or a=-2b. Plugging back x, it becomes 2^{x+2}=3^{x+2} or 2^x=(-2)3^x, where latter is ... 4 Yes, there's a basic rule that you're missing, and it's a pretty useful one. When you have \frac{N^a}{N^b}, you can simplify to N^{a-b}. In your case, a = \frac{2}{3} and b = 1, so the final exponent is \frac{2}{3} - 1 = -\frac{1}{3}. If you're wondering why the rule is true, it's probably easiest to see when a and b are integers. If you ... 3 since \sqrt { 2-\sqrt { 3 } } =\frac { 1 }{ \sqrt { 2+\sqrt { 3 } } }  t=\left( \sqrt { 2-\sqrt { 3 } } \right) ^{ x } you should solve only this eqv. t+\frac { 1 }{ t } =2 3 There is no real solution for the first question 1^x = 2 Not even a complex number. For the second part: Let x = a+ib and e^{(2n+1)i\pi} = -1 , n \in \ Z  so, 2^{a+ib} = 2e^{(2n+1)i\pi} 2^{a-1 + ib} = e^{(2n+1)i\pi} Take log and you will get, (a-1+ib)\log2 = (2n+1)i\pi Equating real and imaginary part, a = 1, b = \frac{(2n+1)\pi}{\log2} ... 3 Other answers have explained why \sqrt{i} is a complex number, and shown how to compute it using complex exponentials, but you can also compute it directly. If you want to find the complex number a + bi, where a and b are real numbers, such that \sqrt{i} = a + bi, then square both sides and solve$$i = (a + bi)^2.$$Expanding the right-hand side, ... 3 For x>0\land x\ne1,$$0<x(x-1)^2\implies x^2<\frac{x+x^3}2.$$Then by monotonicity and convexity of the exponential$$2^{x^2}<2^{\frac{x+x^3}2}<\frac12\left(2^x+2^{x^3}\right).$$For x<0,$$\dfrac12\left(2^x+2^{x^3}\right)<1<2^{x^2}.$$Two cases remain, x=0 and x=1, that happen to be solutions. 3 HINT:$$2^x(1+2^{x^3-x})=2^{x^2+1}\iff 1+2^{x^3-x}=2^{x^2-x+1} If $x^3-x>0,1+2^{x^3-x}>1$ is odd unlike $2^{x^2-x+1}$ as $x^2-x+1>0$ for real $x$ So, $x^3-x$ must be $0$ and consequently, $2=2^{x^2-x+1}\implies 1=x^2-x+1$

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