9
$\begingroup$

Given a number n, how would I convert this number into a logarithmic scale?

My logarithmic scale would range from 0 to 255 (I'm working with RGB colours), and I would expect values of n from 1 to 1,000,000.

Apologies if this is an easy question.

$\endgroup$
11
$\begingroup$

let $f(n)$ be the log function of n.

So a general log function can be written as $f(n)=k\log n +c$

where $k$ and $c$ are constants

$\implies 255=k\log {10^6}+c$

and $0=k\log {1}+c$

here is to the base 10

so $c=0$ and $k=255/6$

or here you can use any base and different values of k

so that $f(n)=\frac {255 \log n}{6}$ if the base is 10.

$\endgroup$
  • 2
    $\begingroup$ Thanks, that's great. How would I approach the problem if I expected values of n that ranged from 0 to 1,000,000 (since log 0 is undefined)? $\endgroup$ – James Williams Oct 12 '14 at 15:33
  • 1
    $\begingroup$ Then you can redifine your function as $f(n)=klog(p+n) + c (p=10^6) $ Here you can get values of k and c by substituting 0 and 255 in the place of $f(n)$ and $n=0, 10^6$ respectively. Don't forget to upvote and if satisfied mark as correct answer. $\endgroup$ – Jasser Oct 13 '14 at 6:36
  • 1
    $\begingroup$ What is p? And how can a function have p = 10^6 in it? $\endgroup$ – Ian Warburton May 11 '15 at 15:52
  • 1
    $\begingroup$ p is just a number so that the value of log is defined even for n=0 case. U can take it as any number such that your function is always defined. Moreover if u do this there is nothing wrong going on because the function would still be logarithmic. @Boumbles $\endgroup$ – Jasser Sep 11 '15 at 15:58
  • 1
    $\begingroup$ @IanWarburton sorry to respond so so late.... thanks to Boumbies' I saw your comment. Are the ranges of logx and log(x+c) equal where c is any real constant. If yes, then taking p=10^6 should not mind u because the function would still be logarithmic. $\endgroup$ – Jasser Sep 11 '15 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.