I am trying to fully understand EV (exposure values) as used in photography. This is a system intended to simplify camera shutter speed/aperture combinations using a single value instead of two.

EV calculation using log base 2 EV=log ((N²/t) where N = camera’s f-number setting and t = shutter time in seconds. This is the classic method.

Alterative using base 10 EV= 3.32 log N2/t

Both equations yield approximately the same answer. What I am asking is the significance of the value 3.32. Is this a conversion factor log base 2 to log base 10?

4 Answers 4


Base conversion of logs is
log(x) / log(new_base) = log<new_base>(x)

Converting between base 2 and base 10:
log2(x) / log2(10) = log2(x) / 3.3219 = log10(x)
Rearranging, log2(x) = 3.3219 log10(x)

In case you're wondering how I got log2(10), I used natural logs to do the conversion, thus
ln(10) / ln(2) = 3.3219

To make the calculation easier for your calculator you can use any arbitrary base and convert to base 2 using the above technique. Thus, your original equation can be calculated as follows using either base 10 or natural log functions commonly found on calculators. No need to memorize the 3.32 conversion constant and shows by observation that you are using a base 2 log.
EV = log2(N2/t) = logb(N2/t) / logb(2)
Where logb can be any base logarithm.

e.g., EV = log10(N2/t) / log10(2)


Yes. 23.32 is approximately 10. Close enough that actual focal lengths, aperture sizes, and shutter timings will have more effect. Not to mention the differences between f-stop and t-stop.


Is this a conversion factor log base 2 to log base 10?

Yes, approximately speaking.

If you require more precision, to sixteen significant digits the conversion factor between log2 to log10 is 3.321928094887362

But, as you are well aware, stated/targeted aperture values versus actual aperture values, as well as stated/targeted exposure times versus actual exposure times are less precise than using 3.32 as your conversion factor.


What I am asking is the significance of the value 3.32. Is this a conversion factor log base 2 to log base 10?

EV = log2(value) = log10(value)/log10(2). Yes, dividing by log10(2) is the conversion to log base 2.

Log10(2) = 0.3, and dividing by 0.3 is same as multiplying by 1/0.3 = 3.33, so either works. And dividing by log(2) also works.

Do NOT use 0.3 or 3.33 in a programming language like Excel or Javascript, because programming languages use log base e instead of log10, and their log(2) is NOT 0.3. So then use / log(2).

  • 0.3 is only an approximation to log10(2), it is not exact. Feb 2 at 4:37
  • So use log10(2), but it is 0.30103, very close to 0.3. Seems close enough for most EV work.
    – WayneF
    Feb 2 at 11:01
  • Sure it's close enough, but your statement is factually incorrect and I'm compelled to point it out: xkcd.com/386 . The question itself had a closer approximation. Feb 2 at 13:02

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