Calculating bits of RAM

You're looking for a logarithm, a base-two logarithm specifically. Logarithms do the opposite of exponentiation, so if b^x = y, then x = log_b y. 2⁴ = 16, so log₂ 16 = 4.

First, you'll need to figure out how many bytes you have. If your number is in kilobytes, multiply by 2¹⁰. For megabytes, 2²⁰, for gigabytes 2³⁰, et cetera. As you can see, I'm using the powers-of-1024 definitions of these units rather the powers-of-1000 definitions, so one kilobyte here is 1024 bytes. The unambiguous name for 1024 bytes is kibibyte. Anyway, 512 MB is equal to 512 • 2²⁰ = 536870912 bytes.

Now you'll need a scientific calculator. I like Wolfram Alpha, which lets you do base-two logarithms with the log2 function. log2(536870912) produces 29, which makes sense, considering that 512 MB is half of 1 GB, so it takes one less power of two. You can use pretty much any operator imaginable in a Wolfram Alpha expression, so log2(512 * 10^20) works too.

If you get a number with a decimal part, round up. For example, you'd need three bits to address five bytes of RAM, though log2(5) is roughly 2.32.

Additionally to what Ben said, I recommend first doing the logarithm of your number without units

log₂512 = 9

And then take units into consideration: sum 10 for kibibytes, 20 for mebibytes, 30 for gibibytes, ...

9 + 20 = 29

And that's it. No need to calculate huge numbers. That's because logarithms have the following properties:

logₙ(a × b) = logₙ(a) +  logₙ(b)
logₙ(aᵇ) = b × logₙ(a)

Therefore,

log₂(512 × 2²⁰) = log₂(512) + 20

However, if you already know log₂(1 GiB) = 30,

log₂(512 MiB) = log₂(1 GiB / 2) = log₂(1 GiB) - log₂(2) = 30 - 1 = 29