Is there a geometric definition of the logarithm function that is non-kinematic and does not involve an infinite procedure?

With base 10, for example. I'm asking for definition, not construction method.

(Napier's 1619 definition was kinematic in nature, as described in this post:

Motivation for Napier's Logarithms).

  • $\begingroup$ I'm a bit confused by your question "Geometric definition, but no construction". They feel contradicting to me. The classical construction problems have the requirements to be created by compass and ruler. With these tools you can perform additions, multiplications, divisions, exponentiations and standard square roots. Using that $\log (1-x) = -\sum_{k=1}^\infty x^n/n$. Theoretically this can be constructed, so values for the logarithm can be geometrically derived/approximated. $\endgroup$ – Jasper Jun 2 '16 at 10:06
  • $\begingroup$ @ Jasper: definition is different from construction, e.g. think about pi. $\endgroup$ – exp8j Jun 2 '16 at 10:10
  • $\begingroup$ @ Jasper: The series expansion is indeed a geometric definition, but not what i'm looking for as it involves limits, so i change the question adding the word "finite". $\endgroup$ – exp8j Jun 2 '16 at 10:18
  • $\begingroup$ You can use the following result as a geometric interpretation of the logarithm: $\log(a)$ is the area under the curve $\frac{1}{x}$ from $x=1$ to $x=a$ if $a\geq1$? If $0< a < 1$ you can use $-\log(\frac{1}{a})$. I don't think this is a real definition though, as it is a result of the definition. $\endgroup$ – Jasper Jun 2 '16 at 10:18
  • $\begingroup$ @ Jasper: i agree, the problem is to find a definition. $\endgroup$ – exp8j Jun 2 '16 at 10:23

If you have a exponential equation and plot the function on a 2D surface, Let, $y=f(x)=b^x$

Then,if you plot the inverse function of the above equation $f^{-1}(x)$,then you get a curve,that curve describe the logarithm.

  • $\begingroup$ @ Hailey: yes, but how do i plot it?. $\endgroup$ – exp8j Jun 2 '16 at 7:51
  • $\begingroup$ Let,b=10.so,take x=0,1,2,3,..and you get curve for $10^{x}$.Now now determine for which $x$ value $10^{x} = 0$ that will give (x,0)point for log(x).And for the values less than x it will give you -ve $y$ values.And for greater value than x give you +ve $y$ value.and so on $\endgroup$ – Hailey Jun 2 '16 at 7:59
  • $\begingroup$ @ Hailey: apart from 10^x =0 being impossible, taking x=0,1,2,3.. isn't sufficient for plotting the curve. $\endgroup$ – exp8j Jun 2 '16 at 8:52
  • $\begingroup$ There was a mistake in the comment <br> Listen $10^{-2},10^{-1},10^{0},10^{1},10^{2},10^{3}$...gives you exponential curve. And to determine the log(x) value,for which value $10^{x}=1$ that gives you (x,0) ,similarly,for which value of x you will get $10^{x}=10,100,1000$. I take these value as this value will give you the integral solution for x $\endgroup$ – Hailey Jun 2 '16 at 8:56
  • $\begingroup$ I changed the original question, to non-kinematic definition. $\endgroup$ – exp8j Jun 2 '16 at 9:56

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