# from where was the logarithmic constant pulled from?

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from where was the logarithmic constant pulled from?

In: Mathematics

The derivative of an exponential is an exponential as well, so there will also be an exponential whose derivative is itself, and that is e^x

And it turns out because of that it also has lots of other mathematical proprieties and appears very often in maths

I assume you mean “e” the exponential constant? It’s not pulled from somewhere, it naturally arises from a lot of exponential or logarithmic relationships.

Easy example is compound interest. If you invest \$1 at a bank that pays 100% annual interest, how much you end up with depends on how frequently the bank pays out. The formula is “Total money = (1 + 1/n) ^ n” where “n” is how frequently you are paid. The more frequently you’re paid, the more money you end up with:

annual = (1 + 1/1) ^ 1 = \$2

biannual = (1 + 1/2) ^ 2 = \$2.25

quarterly = (1 + 1/4) ^ 4 = \$2.44

monthly = (1 + 1/12) ^ 12 = \$2.61

weekly = (1 + 1/52) ^ 52 = \$2.69

daily = (1 + 1/365) ^ 365 = \$2.71

and so on. If you keep shrinking the time between payments until the gap is infinitesimally small (e.g. take the limit), you will get \$2.71828…. which is “e.”

Expanding on the derivative answer –

If we have an exponential function f(x)=a^x (for a>1), that gives us an exponential curve. The curve shrinks down to zero as we look at x on the left side of the graph (as x -> negative infinity), at x=0, the result is 1, at x=1, the result is a, and it continues to grow ever faster as we look to the right on the graph.

Now, this function is continuous and it has a slope (how fast it is growing) for each x. This is called the “derivative” of the function.

For any exponential function, with some limit math, you can show that the derivative f'(x) is equal to a^x times some constant c. (f'(x) = c a^x ). And since a^0 = 1 (remember that a>1), you can also show that that constant is equal to the slope at x=0 for the function. So you get f'(x) = f'(0) a^x.

Well, the question naturally arises = “for which a is f'(0) = 1”? Which is the same thing as asking “for which a is f'(x) = a^x” or (as SuperReddit 578 said) “for which function is f'(x) = f(x)”.

Through some more limit math, you can show that there is such a constant, and its value is right around 2.718…. It’s not rational, so the decimal expansion goes on forever, but a few digits is usually all we need. We called that value “e”, and called the logarithm with respect to that base “the natural log” which we abbreviate as “ln” (log naturale).

Going back to the equations above, it turns out that for any f(x) = a^x, f'(x) = ln(a) a^x. Since ln(e) = 1, we get back to f'(x) = e^x for f(x) = e^x.