# Tag Info

5

Well, $e^{\pi i/2}=i$, so $\log i$ is $\frac{\pi}{2}i$ -- plus or minus some multiple of $2\pi i$, but let us work with $\frac\pi2i$ as the principal value to begin with. Then $$\log(\frac\pi2i) = \log(\frac\pi2) + \log i = \log(\frac\pi2) + \frac\pi2i$$ This matches Google's result, since $\log(\frac\pi2)\approx 0.452$ and $\frac\pi2\approx1.57$. In is ...

4

Here is a picture that matches the description of the path of integration: Note that we are taking the positive real axis as the branch cut for $\log(-x)$, and we have taken the branch of $\log(-x)$ whose argument goes from $-\pi$ to $\pi$. The red and blue lines should be infinitesimally above and below the positive real axis.

4

Note that $\ln (x+2)=\ln 2+\ln \left(1+\cfrac x2\right)$ will readily give you the same as wolfram. If you look carefully at the first of the series representations in the link you provided, you will see that it is the same as your expression.

3

OK I tried it. Now what? Is this different from yours?

3

Hint: Use strengthenings of Bertrand's Postulate, in particular Nagura's result that for $n\ge 25$ there is always a prime between $n$ and $n\left(1+\frac{1}{5}\right)$. We can use this to show that unless $k$ is very small, there are always at least $4$ distinct primes $p$ that satisfy $\lfloor \log p \rfloor=k$.

3

As an alternative to Aniket's answer, we can carefully specify what we mean by $d\ln(x)$ and still get the same result. I'll work from right to left. Let $y(x) = \ln(x)$, whence $x(y) = e^y$. We compute the derivative with respect to $y$ $$\frac{d}{dy} \ln(f(x)) = \frac{dx}{dy} \frac{d}{dx} \ln(f(x)) = \frac{dx}{dy} \frac{1}{f(x)} \frac{d}{dx} f(x).$$ We ...

2

$\frac{x}{f(x)}\frac{d(f(x))}{dx}=\frac{x}{f(x)}\cdot f'(x)=x\cdot \frac{f'(x)}{f(x)}$ $=\large\frac{\frac{f'(x)}{f(x)}}{\frac{1}{x}}=\large\frac{\frac{d}{dx}[\ln(f(x))]}{\frac{d}{dx}[\ln x]}=\frac{d\ln{f(x)}}{d\ln{x}}$ Hope this helps.

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

Hint: use the fact that $$10=(10^{-1})^{-1}$$

2

Using the following rules: 1) $\log_{a}(b)=\frac{\ln(b)}{\ln(a)}$; 2) $\ln\left(\frac{1}{a}\right)=-\ln(a)$; 3) $\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$. ...

2

(1) $a=(5+4\sqrt{2})-(4\sqrt{2}-5)=10$ (2) $b=\sqrt{25^{\log_5{8}}+49^{\log_7{6}}}=\sqrt{8^2+6^2}=10$ (3) $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}=(\sqrt{2}+1)-\frac{1}{\sqrt{2}+1}=(\sqrt{2}+1)-(\sqrt{2}-1)=2$, $c=8+6-14=0$

2

$$a^2=57+40\sqrt2+57-40\sqrt2-2\sqrt{57^2-40^2\cdot2}=2\cdot57-2\sqrt49=100$$ $$a=10$$ $$b=\sqrt{5^{2\log_5 8}+7^{2\log_7 6}}=\sqrt{8^2+6^2}=\sqrt100=10$$ $$x=a-\frac1a,\quad a=(7+5\sqrt2)^{\frac13}$$ $$x^3+3x-14=\left(a^3-3a+\frac3a-\frac1{a^3}\right)+3\left(a-\frac1a\right)-14\\ ... 2 \log(−x) for x>0 would be a logarithm of negative number, and it doesn't exist. 2 We have in general$$y(x)=\log_{f(x)}e\implies y(x)\ln\left(f(x)\right)=1\implies y'(x)=-\frac{f'(x)}{f(x)\ln^2(f(x))} \tag 1$$For f(x)=1/x, (1) gives$$y'(x)=\frac{1}{x\ln^2(x)}$$while for f(x)=\ln(x), (1) gives$$y'(x)=-\frac{1}{x\ln(x)\ln^2(\ln(x))}$$1 There is more to what they specified than you've reproduced here. \log(-x) is not necessarily undefined when the values are allowed to be complex. In fact, if x = re^{i\theta}, then -x = re^{i(\theta + \pi)}. Therefore we can define \log(-x) = \log(r) + i(\theta + \pi). The problem is, -x = re^{i(\theta + (2n+1)\pi)} as well, for any integer n. ... 1 Since \log_{i} n = \frac {\log n} {\log i}, we have$$S_n = \sum_{i = 2}^{n} \frac {\log n} {\log i} = \log n \sum_{i = 2}^{n} \frac {1} {\log i}$$and we have by Euler-McLaurin summation formula$$\sum_{i = 2}^{n} \frac {1} {\log i} = \int_{2}^{n} \frac {\text {d} x} {\log x} + \log \sqrt {2n} + O \left(\frac {1} {\log n}\right),$$which leads us to$$S_n ...

1

Since the base of exponential is 10, the correct value for $x$ will be $-1$ and $-10$.In other words, if $\sqrt{\log(-x)}=\log{-x}$ then the equation has solution. Check by putting in the value.

1

It is a godo approach to split into cases depending on the sign of $x$. Clearly $x=0$ is not a solution, so you just consider (as you do) $x >0$ and $x<0$. If $x >0$, then $\log(-x)$ isn't defined because $-x<0$, so there are not solutions where $x > 0$. For $x<0$ you find (as you do) $\sqrt{\log(-x)} = \log(-x)$. The solution to the ...

1

Complex logarithm is not a universally defined function. $$re^{i\theta} = re^{i\theta + 2\pi k i}$$ for any $k \in \mathbb Z$, so one has to specify which $k$ is preferred before using the complex logarithm. Even after specifying $k$, there is always some ray from the origin in the complex plane on which the formulation of $\log$ is undefined. In the case ...

1

Let $\log(e^{5i})=z$. This means that $e^z=e^{5i}$. Now, if $z=a+ib$ this means: $$e^{a+ib}=e^{5i} \iff e^ae^{ib}=e^{5i}$$ so we have : $$e^a=1 \Rightarrow a=0$$ and $$b=5+2k\pi$$ because the exponential is a periodic function with period $2i\pi$. This means that his inverse function (the logarithm) is multi-valued, and you can have a single value, or ...

1

Apply the logarithm rule $\color{blue}{\large \log_{a^m}(b^n)=\frac{n}{m}\log_a(b)}$, hence $$\frac{1}{2}\log_{10^{-1}}10-\log_{10^{-1}}10^2$$ $$=-\frac{1}{2}\log_{10}10-2(-1)\log_{10}10$$ $$=-\frac{1}{2}+2=\color{red}{\frac{3}{2}}$$

1

$x^{\log_3 x} = 81$. Take $\log_3$ of both sides: $\log_3 x \cdot \log_3 x = \log_3 81 = 4$, so $(\log_3 x)^2 = 4$. You should be able to finish it now.

1

$\Delta x / x_0$ is a constant ($1/x_0$) multiplied by $\Delta x$, which doesn't change the answer you get when you consider a "small quantity" $\Delta x$ or $t = \Delta x / x_0$. However if you put $t = x_0 / \Delta x$ this is not proportional to $\Delta x$ (it is proportional to $1/ \Delta x$) so you have to treat the limit differently (i.e., letting \$t ...

1

As far as I can tell, the table-maker's dilemma as described in the sources listed in the comments Wikipedia Table-Maker's dilemma and Intro to TMD refers to computing in advance, given the IEEE rounding mode (nearest, truncate, floor, ceiling), the operation to be performed, and the range of arguments, the greatest number of significant figures n in order ...

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