# Tag Info

4

Let's say we have two fractions in an equality: $$\frac{a}{b}=\frac{c}{d}$$ where $b\neq0\neq d$. We can flip them over, and say that $$\frac{b}{a}=\frac{d}{c}$$ where $a\neq0\neq c$. Let's apply that here. First, we distribute out the $x$: $$\left(\frac{1}{2}\right)^x=\frac{1^x}{2^x}$$ Now we substitute this back in: $$\frac{1^x}{2^x}=\frac{1}{7}$$ ...

4

$A$ is an order-$2$ permutation matrix, hence: $$e^{A}=\sum_{n\geq 0}\frac{A^n}{n!} = A\sum_{n\geq 0}\frac{I}{(2n+1)!}+I\sum_{n\geq 0}\frac{I}{(2n)!} = \sinh(1) A + \cosh(1) I = \begin{pmatrix}\cosh(1)&\sinh(1)\\\sinh(1)&\cosh(1)\end{pmatrix}=\frac{1}{2e}\begin{pmatrix}e^2+1&e^2-1\\e^2-1&e^2+1\end{pmatrix}.$$

4

Hint You may try with the Sophie Germain identity. As $(10^4+324)=(10^4+4\times3^4)$

4

Here is a dirty but quick way as a rule of thumb: we have $$\bigg( \frac{5}{a^{4}} \bigg)^{-3} = \bigg( \frac{a^{4}}{5} \bigg)^{3} = \frac{a^{12}}{125}$$ I use "dirty" here because the number $a$ should be $\neq 0$.

4

You are right: $i$ is another solution of the equation $z^3=-i$. There is also a third one, $\dfrac{1+i\sqrt{3}}{2}$. However, this being a multiple choice question, your choices are constrained to the options given. Only one of the three roots was among the options.

4

A CAS tells me $$\gcd(2^{83}+1,3^{83}+1)=499$$ So the claim is false. It happens to be true for many primes $p$, but that is just because of the abundance of possible divisors available for large numbers like $2^{p}+1$ and $3^{p}+1$, and the consequent unlikelihood of an overlap. And yet it's not impossible to have overlap, as with $p=83$. (Looks like ...

3

If I understand the question, then what you wrote is correct. You can verify properties of exponentiation on corresponding Wikipedia page, for example. In particular, \begin{align} \big(\,b^m\,\big)^n &= b^{m\cdot n}& \text{ and }& & b^{-n} &= \dfrac{1}{b^n} \end{align}

3

The two standard ways are: 10^X and 10 ** X. Either one of those should be almost universally recognized.

3

As it seems to be a multiple choice question, we can find the answer without much computing: all factors have the form $$(2n)^4+324=4(4n^4+81).$$ Since there are as many factors in the numerator as in the denominator, the $4$s cancel out, which results in a fraction with odd numerator and denominator, hence this fraction must simplify to an odd number. ...

3

Hint. By the chain rule, you rather have $$(\tan^2(x^4))'=2\times 4x^3\times \sec^2(x^4)\times\tan(x^4)$$ and $$(\sec^3(x^5))'=3\times 5x^4\times \sec(x^5)\times\tan (x^5)\times\sec^2(x^5)$$

2

$(a+bi)^{(c+di)}=e^{(c+di)\ln(a+bi)}$ and $\ln(a+bi)=\ln\left | a+bi \right |+i\arg(a+bi)$ and $e^{m+ni}=e^m(\cos(n)+i\sin(n)).$

2

Note that if $n$ is a positive integer then $$\left(\frac{4}{3}\right)^n =\left(1+\frac{1}{3}\right)^n\ge 1+\frac{n}{3}$$ (Bernoulli Inequality).

2

Hint: Instead of $\cos$ and $\sin$, think $\cosh$ and $\sinh$. These hyperbolic trig functions have almost the same expansion, except each term is positive.

2

These three numbers have the same ordering as their tenth roots, which are $2^{12}$, $3^8$, and $10^3$; all of these are easily calculated by hand. The hardest is $3^8$, and it’s pretty easy: $$3^8=9^4=81^2=6561\;,$$ all of which was done in my head.

2

This is correct, next step is to take the 3-log of both sides: $$\log_3(x^{\log_3 x}) = (\log_3 x)(\log_3 x) = \log_3 81 = 4$$ So $\log_3 x = \pm 2$ which means that $x = 3^2 = 9$ or $x=3^{-2} = 1/9$. As for the first step that is correct $\log_33^{(\log_3x)^2} = (\log_3x)^2\log_33 = (\log_3x)^2 = \log_3(x^{\log_3 x})$, so the terms on the left hand ...

2

The sequence of partial sums is on the OEIS, but no simple formula is listed there, which makes me think no such formula is known.

2

The correct order of the operations is: first calculate the powers than add the results. So in you case: $$-5-8^2=-5-(8^2)=-5-64=-69$$

2

Using the following identity: $$a^4 + 4\cdot 3^4 = (a^2 + 2 \cdot 3^2 - 2\cdot 3\cdot a)(a^2 + 2 \cdot 3^2 + 2\cdot 3\cdot a) = (a(a-6) + 18)(a(a+6)+18)$$ Most of the terms cancel out and you are left with: $$\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = 373$$ As KprimeX mentioned, this flows from the Sophie Germain Identity.

2

You have: $$\left(\frac{5}{a^4}\right)^{-3}=\left(\frac{5^{-3}}{a^{-12}}\right)=\left(\frac{a^{12}}{5^3}\right)=\frac{a^{12}}{125}$$

2

The general answer can be given as follows . $i^3=-i$ so we write $x^3=-i$ now we want three solutions where one is $i$ so for solutions we can write $cos{\frac{2kπ+\theta}{n}}+isin{\frac{2kπ+\theta}{n}}$ where $k=(0,1..,n-1)$ so here $n=3$ so $k=(0,1,2)$ now plug in the values and get the answer. Here $\theta=3π/2 ,n=3$ hope you can do the rest.

2

No. Suppose that $f(a,b)=d$ is such a function. Then $f(2,2)=4$ since for $c=1$ you get $a^{(b^1)}=2^{(2^1)}=2^2=4$. Now notice that for $c=3$ you get $2^{(2^3)}=2^8\neq 4^3=2^6$.

1

$$d=a^{b^c/c}$$ so no, because $b^c/c$ depends on $c$. For example, when $c=1$, $b^c/c=b$, when $c=-1$, then $b^c/c=-1/b$. Since $x\mapsto a^x$ is one-to-one, this means that in general, $d$ requires all three variable values as input.

1

It's a one-liner in Maple: convert(convert(2^1000, base, 10),+); You could look up OEIS sequence A001370. Or you could just ask Wolfram Alpha. But if you're asking for a way of doing it by hand, I very much doubt that there is any.

1

As you said, $x=1$ is solution. Moreover, if $f(x)=2x+2^x$,$$f'(x)=2+\ln(2)2^x>0$$ and thus, $f$ is injective. Therefore, $x=1$ is the unique solution.

1

Some people are asserting that there is no algebraic solution, but they seem to not have tried very hard. To start, $$2x + 2^x = 4 \\ 2^{x-1} = 2 - x \\ e^{(x-1)\log 2} = 2 - x \\ 1 = e^{(1-x)\log 2}(2-x) \\ e^{\log 2} = e^{(2-x)\log 2}(2-x) \\ e^{\log 2}\log 2 = e^{(2-x)\log 2}(2-x)\log 2.$$ The left hand side is positive, and so the right must be too. ...

1

$$3 = 2^{\log_23}$$ Therefore $$3^n = (2^{\log_23})^n=2^{n\log_23}$$

1

Assume that $\exp(A)$ has no eigenvalues in $(-\infty,0]$ and let $\log(.)$ be the principal logarithm. Then we can define $(\exp(A))^{\alpha}=\exp(\alpha \log(e^A))$. For the sake of simplicity, assume that $A=diag(\lambda_j)$; then $\log(e^A)=diag(\log(e^{\lambda_j}))=diag(\lambda_j+2k_ji\pi)$ where $k_j\in\mathbb{Z}$. Finally ...

1

This relates to the deeper questions of where does the iterates of $a^z$ converge in the complex plane for infinite exponential towers at a fixed point $A$ such that $a^A=A$ and where is the onset for period $n$ behavior. Let $\Delta z$ be an infinitesimal. Since $a^A=A$, $a^{A+\Delta z}=A a^{\Delta z}=A + Ln A \ \! {\Delta z}$. Therefore $A+\Delta z ... 1 Since $$-i=0+(-i)=\cos(3\pi/2)+\sin(3\pi/2)i$$ then you'll have one of the below versions for$z$: $$z=\cos\left(\frac{2k\pi+3\pi/2}{3}\right)+i\sin\left(\frac{2k\pi+3\pi/2}{3}\right),~~k=0,1,2$$ 1 We have $$n = (2^{2^{2^{15}}})^2=2^{(2^{2^{15}}+1)}$$ The$+1$is negligible. If you take one log, you have $$\log_2(n)=2^{2^{15}}\\\log_2(\log_2(n))=2^{15}\\$$ A third one gets you to$15$, then a little less than$4$, then a little less than$2$, then less than$1\$. So it takes six applications of the log function.

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