Questions tagged [logarithms]
Questions related to real and complex logarithms.
10,256
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Alternative notation for exponents, logs and roots?
If we have
$$ x^y = z $$
then we know that
$$ \sqrt[y]{z} = x $$
and
$$ \log_x{z} = y .$$
As a visually-oriented person I have often been dismayed that the symbols for these three operators ...
- 2,713
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votes
27
answers
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Unexpected examples of natural logarithm
Quite often, mathematics students become surprised by the fact that for a mathematician, the term “logarithm” and the expression $\log$ nearly always mean natural logarithm instead of the common ...
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votes
10
answers
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The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$) [duplicate]
In general, the indefinite integral of $x^n$ has power $n+1$. This is the standard power rule. Why does it "break" for $n=-1$? In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ ...
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answers
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Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$
Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find ...
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21
answers
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How do you explain the concept of logarithm to a five year old?
Okay, I understand that it cannot be explained to a 5 year old. But how do you explain the logarithm to primary school students?
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votes
11
answers
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Demystify integration of $\int \frac{1}{x} \mathrm dx$
I've learned in my analysis class, that
$$ \int \frac{1}{x} \mathrm dx = \ln(x). $$
I can live with that, and it's what I use when solving equations like that.
But how can I solve this, without ...
- 1,051
85
votes
3
answers
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What algorithm is used by computers to calculate logarithms?
I would like to know how logarithms are calculated by computers. The GNU C library, for example, uses a call to the fyl2x() assembler instruction, which means that ...
- 4,602
83
votes
2
answers
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Ramanujan log-trigonometric integrals
I discovered the following conjectured identity numerically while studying a family of related integrals.
Let's set
$$
R^{+}:= \frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}
...
- 121k
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votes
6
answers
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What's so "natural" about the base of natural logarithms?
There are so many available bases. Why is the strange number $e$ preferred over all else?
Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
user218
74
votes
5
answers
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A new imaginary number? $x^c = -x$
Being young, I don't have much experience with imaginary numbers outside of the basic usages of $i$. As I was sitting in my high school math class doing logs, I had an idea of something that would ...
- 853
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votes
5
answers
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Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$
Is there a closed form for the integral
$$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$
I do not have a strong reason to be sure it exists, but I ...
- 1,249
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votes
4
answers
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Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
I'm interested in integrals of the form
$$I(a,b)=\int_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx,\color{#808080}{\text{ for }a>0,\,b>0}\...
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Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
Please help me to find a closed form for the following integral:
$$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$
I was told it could be calculated in a ...
- 13.2k
68
votes
4
answers
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Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$
One of fundamental inequalities on logarithm is:
$$ 1 - \frac1x \leq \log x \leq x-1 \quad\text{for all $x > 0$},$$
which you may prefer write in the form of
$$ \frac{x}{1+x} \leq \log{(1+x)} \leq ...
- 1,908
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votes
12
answers
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,...
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