Is there some special property of '$e$' which makes it suitable to be used as a base for logarithms?  Moreover, does the natural logarithm possess some advantage over the common logarithm? I don't understand why there is a need to choose an irrational number, '$e$', for a base. Isn't it much simpler to use 10 as a base?
98951
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?
22919
What does the ratio of the circumference of a circle to its diameter have to do with the base of the natural logarithm and $\sqrt{-1}$?
67326
As the title states, I need to be able to calculate logs (base $10$) on paper without a calculator. For example, how would I calculate $\log(25)$?
96262
Could you help me fill in a gap in my understanding of maths?  As long as the exponent is rational, I can decompose it into something that always works with primitives I know.  Say, I see a power: $x^{-{a\over b}}$ for natural a,b.   Using the basic rules:   $$\begin {align} x^{-a} =&amp; {1 \over x^a} \\ x^{1\over b} =&amp; \sqrt[b]{x} \\ ({x^a})^b =&amp; {x^{a b}} \end{align}$$  I can always decompose rational exponents into $ x^{{1\over b} \cdot -a} = \sqrt[b]{x} ^ {-a} = {1 \over \sqrt[b]{x}^a }$- and integer degree roots are something well within my grasp. Integer root is always just simply a number of multiplications, $\sqrt[a]{x}$ is just some $y \cdot y \cdot y \cdot ... \cdot y$, as many $y$'s as $a$ says.   If the exponent is irrational, it's trickier, but I can always take $e^{\pi} \approx e^{314159265358979323 \over 100000000000000000}$ and take more $\pi$ digits if needed, and even though the number of underlying multiplications becomes ridiculous, that's still something I can understand.  But I'm completely at loss how to understand stuff like $x^{i\pi \over 4}$. I just can't perform an imaginary number of multiplications. I know how to perform conversion of complex numbers between the exponential and $a+ib$ form, but I perform it like a mysterious voodoo magic recipe without ability to wrap my mind around how that kind of exponentiation is supposed to work. I can still apply the old conversion formulas, $x^a \cdot x^b = x^{a+b}$; $({x^a})^b = {x^{a b}};$ and obtain correct results but I just don't understand the underlying mechanism - as $x^n$ is just $n$ multiplications of $x$, how can understand how that works when $n$ is imaginary?
3586
I am trying to gain a more intuitive feeling for the use of logarithms.   So, my questions: What do you use them for? Why were they invented? What are typical situations where one should think: "hey, let's take the $\log$!"?     Thanks for the great comments!      Here a short summary what we have so far:      History:      Logarithms were first published 1614   by John Napier   (mathematician/astronomer) . He needed   them for simplifying the   multiplication of large numbers.      Today’s uses:         In regression analysis: If you expect a predictor variable to follow   a power/exponential law, take the   corresponding logarithm to linearize   the relationship.   In finance to calculate compound interests.   Or more general: to calculate the time variable in growth/decay   functions. (Nuclear decay, biological   growth…)   In computer science to avoid underflow. (Cool trick! But seriously:   32-bit? Take a double ;-)   In the Prime Number Theorem   For handling very large/small numbers (pH, etc.)   Plotting (if your scale gets too large)
64115
Are $\log_1 1$ and $\log_0 0$ indeterminate forms?  Whenever I ask someone about these indeterminate forms, they deny by saying either $\log$ is neither defined at base $0$ nor at base $1$, or they say $\log$ is a function so these must not be included in fundamental indeterminates.  But, we know division by zero is not defined, yet $0/0$ is indeterminate; and many others. And, actually, $\log$ is more a binary-operator that is the inverse operation of power/exponent operator.
36720
I know when evaluated it gives one but I can't figure out how to prove it.  Can anyone help? I believe it requires L'Hopitals rule and taking a natural log but i cannot figure out the exact math.
90289
This question is me trying to understand (again) why there can be no one-to-one correspondence between the sets of natural and real numbers.  The source of confusion is this: if we abstract completely from the properties and operations on numbers and just think of them as strings of digits, it appears to be possible to list all the real numbers between $0$ and $1$ and all positive integers having the same digits in the same (or reverse) order, for example:  $$0.1 \to 1$$  $$0.01 \to 10$$  $$0.0101 \to 1010$$  $$0.110101 \to 101011$$  I use binary expansion for simplicity.  Since irrational numbers have infinite expansions, we would have to consider some kind of 'infinite natural numbers', like:  $$101110001\dots$$  or  $$\dots101110001$$  I think the main (and only) reason why such numbers don't exist is this requirement:   We need the sets of real and natural numbers to be ordered. We need to be able to do arithmetics on real and natural numbers.   Obviously, if we consider any of two types of 'infinite natural numbers' and the usual absolute value, we find that it's impossible to order them, or do arithmetics with them.  The only way is to change the norm (to $p-$adic norm for example).  While with the real numbers it's still possible to order them, and do arithmetics with them when they have infinite amount of digits in their expansion (or at least in principle).     Is my reasoning correct? Are the two requirements (ordering and arithmetics) enough to explain why natural numbers with infinite amount of digits do not exist, while real numbers do?   This confusion is not restricted to me alone, see  for example
68266
I am interested in good integer approximation from below and from above for binary Log(N!).  and  provides only a general idea but not exact values. In other words I need integers A and B so that  A &lt;= Log(N!) &lt;= B
63508
Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.  Here is a simple technique used to approximate square roots by :  For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.  if $n=2^k$ we can use the method above.   For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 }  } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 }  } =\sqrt { 11\frac { 15 }{ 23 }  } \\ \sqrt { 11\frac { 15 }{ 23 }  } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know.
81670
As the natural $\log(x)$ function is the inverse of the exponential $e^x$ and $\log(x +1)$ is the inverse of $e^x - 1$, what it the inverse of $e^x + x$?
71784
It's famously unknown whether the natural log of 2 is rational or not.  How about the natural log of other numbers. Is it known/unknown whether these are rational?  Obviously ln(1) is 0, and ln(2^n) is n*ln(2) (and is thus rational iff ln(2) is rational), but how about other cases?
84830
In the wikipedia article on logarithms, I am clueless about the approach and motivation for the following computations done by Napier (and the mysterious appearance of Euler's number) in this . Please help me understand:       By repeated subtractions Napier   calculated $10^7(1 − 10^{−7})^L$ for L   ranging from 1 to 100. The result for   $L=100$ is approximately $0.99999 = 1  − 10^{-5}$. Napier then calculated the   products of these numbers with $10^7(1  − 10^{−5})^L$ for L from 1 to 50, and   did similarly with $0.9995 ≈ (1 −  10^{−5})^{20}$ and $0.99 ≈ 0.995^{20}$.   These computations, which occupied 20   years, allowed him to give, for any   number ''N'' from 5 to 10 million, the   number ''L'' that solves the equation      $$N=10^7 {(1-10^{-7})}^L$$      In modern notation, the relation to   natural logarithms is:      $$L = \log_{(1-10^{-7})} \!\left(  \frac{N}{10^7} \right) \approx 10^7  \log_{ \frac{1}{e}} \!\left(  \frac{N}{10^7} \right) = -10^7 \log_e  \!\left( \frac{N}{10^7} \right)$$      where the very close approximation   corresponds to the observation that      $${(1-10^{-7})}^{10^7} \approx  \frac{1}{e}. $$   I understand the computations, but I don't know what's going on.
83091
Why is it such a good idea to plot the logarithm of the form factor vs $Q^2$ in . It seems arbitrary to me.
11871
I've heard a lot about this number $e$. Why is it so important? How does it fit into the 'bigger picture' of mathematics? How is it calculated and used?
50431
Why the log? Is it there to make the growth of the function slower?  As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\infty)$ to $(-\infty,\infty)$ (For a particle emitted along the beam axis after collision $\theta = 0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)
100726
The product of the two roots of $\sqrt{2014}x^{\log_{2014} x}=x^{2014}$ is an integer. Find its units digit.   I'm quite unable to solve the problem given. I have no idea how to deal with that $\sqrt{2014}$ term nor the logarithms in the exponent. Is it even necessary to solve for each individual root or is there some other way to progress through the problem?  I'm really just a novice so I'd appreciate if anyone could show me the process or processes, if multiple. Thank you very much.