Why does the EV formula use 3.32 with base-10 logarithms?

Question

I’m trying to understand the math behind exposure value (EV). The standard formula is EV = log2(N² / t), where N is the f-number and t is shutter time in seconds. I’ve also seen it written using base-10 logs as EV ≈ 3.32 × log10(N² / t). What does the 3.32 represent? Is it just the conversion factor between base-2 and base-10 logarithms?

ev aperture metering shutter-speed exposure

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Photography Stack Exchange contributor

4y ago

user98537 · Accepted Answer

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

[edit]
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) = log_b(N²/t) / log_b(2)
Where log_b can be any base logarithm.

e.g., EV = log₁₀(N²/t) / log₁₀(2)

Why does the EV formula use 3.32 with base-10 logarithms?

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