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Why we take (base $e$) and (base 2) in natural and discrete logarithm, respectively?

I do not understand the difference between these two concepts.

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    $\begingroup$ Reading is often a good starting point...en.m.wikipedia.org/wiki/Discrete_logarithm ...this does not help? $\endgroup$
    – DrLecter
    Commented Mar 24, 2016 at 5:48
  • $\begingroup$ Please explain exactly what you do not understand. This is not the right place to completely describe the discrete logarithm problem. $\endgroup$
    – Maarten Bodewes
    Commented Mar 24, 2016 at 10:03

1 Answer 1

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I will assume you understand modulo operation and the exponentiation.


First let's consider logarithm in $\mathbb{R}$.

You know that if we have $e^x = y$ then $x = \ln y$. The Napierian logarithm return values in $\mathbb{R}$.
You can have the same thing with another base. For example : $2^3 = 8$ and $\log_2 8 = 3$.

The simplified idea of the discrete logarithm is to return only the Integers ($\mathbb{Z}$).

In $\mathbb{R}$, we have $\log_2 5 = 2.321\ldots$. This is not interesting for us. Hard to manipulate... How does this translate into a Integer ?


Lets consider consider the integers where all numbers have to be below a certain prime number $p = 11$.
Lets also consider the only allowed operation is the multiplication : $\times$.

Therefore we have :
$3 \times 4 = 12 \equiv 1 \pmod{11}$
And so on.

This kind of structure is called a group of multiplication modulo the prime p. Its notation is the following $\mathbb{Z}_p^\times$. In this case we have $\mathbb{Z}_{11}^\times$

Let's consider $g=2$ and apply the exponent ($g^n = g \times \ldots \times g$) :
$2^0 \equiv 1$
$2^1 \equiv 2$
$2^2 \equiv 4$
$2^3 \equiv 8$
$2^4 \equiv 2\times 2^3 \equiv 2 \times 8 \equiv 16 \equiv 5 \pmod{11}$
$2^5 \equiv 2\times 2^4 \equiv 10 \pmod{11}$
$2^6 \equiv 2\times 2^5 \equiv 9 \pmod{11}$
$2^7 \equiv 2\times 2^6 \equiv 7 \pmod{11}$
$2^8 \equiv 2\times 2^7 \equiv 3 \pmod{11}$
$2^9 \equiv 2\times 2^8 \equiv 6 \pmod{11}$
$2^{10} \equiv 1 \pmod{11}$

This is called a cyclic group of generator $g$ (here 2) : after a certain number of exponentiations, we have loop.

Now let's move back to our question : $\log_2 5 = ?$
In $\mathbb{Z}_{11}^\times$ as we have $2^4 \equiv 5 \pmod{11}$, therefore in $\mathbb{Z}_{11}^\times$, $\log_2 5 = 4$.

The discrete in discrete logarithm refer to the aspect that we are working in a discrete group and not any real numbers.

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    $\begingroup$ Discrete does not refer to the fact that the exponent is an integer, it refers to a discrete group. $\endgroup$
    – Aleph
    Commented Mar 24, 2016 at 11:01
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    $\begingroup$ Yes but the discrete group implies the later, hence why I think it is easier to remember that way (even though not completely correct) if one does not have a strong mathematical background. $\endgroup$
    – Biv
    Commented Mar 24, 2016 at 12:24
  • $\begingroup$ It is perfect Answer for me. Thanks Biv. Im Studying ECC for make ECC SW Library in Soc Company I Saw Discrete Logarithm Theory but theyre just say equation. Your Answer is very good. I think Discrete Logarithm is adapted for Digital. $\endgroup$
    – pftpmlp
    Commented Mar 24, 2016 at 12:52
  • $\begingroup$ @Biv Sure, but I wouldn't say the converse holds. I would never refer to the logarithm of $b^k$ to base $b$ ($b \in \mathbb R, k \in \mathbb Z)$ as discrete. That said, it was a minor detail and the answer is nice overall. $\endgroup$
    – Aleph
    Commented Mar 24, 2016 at 17:59

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