Question

2 raise to x-5 = 256

Answer 1

To throw in an alternative for fun, you can do it by binary search. This isn't optimal unless you rig the rules, but it shows a method. Start squaring 2 until you get past 512.

2*2 = 4

4*4 = 16

16*16 = 256

256*256 = 65536 , too big

so somewhere between 28 and 216

How about 212 = 28 * 24 which you have as 256 * 16 = 4096 , too big.

Okay, split it again, between 28 too small and 212 too big, 210

210 = 28 * 22 = 256 * 4 = 1024

Still a little bit too big. Next one will get you 29 between 28 and 210.

Best way asymptotically is (unless this has been improved recently) the log using AGM method for logarithm.

That's for finding x where 2x is a number with a gazillion digits.

Best way for small numbers has to do with the circuitry of the calculator.

Going meta, it really depends on the context of the problem. Paper and pencil? In your head? Are you designing a circuit to do this? Is it just this particular question, in which case the easiest way is writing down 9, or what class of variations are you trying to do and what class of methods are you considering and what measurement are you using to decide which is better than another?

Answer 2

2^(x-5)=256

2^(x-5)=2^8

So x-5=8

Therefore x=13