For countries with low denomination currency like UK/US how long will it take for say $50 to be like a $1 today, so things like a $4 beer today would cost $200 in x amount of years.

Well...inflation changes yearly, so it may not be predicted from 2012+.

So first you need an inflation rate, and then it is just algebra from there.

x = years

Inflation rate = .04 or 4% (hypothetically)

$200 = $4*(1+.04)^x <---- compound interest formula (i think you have to use logarithms on these)

Correct me if I'm wrong. I could be.

Given a constant rate of inflation (something that doesn't exist in nature and is very important) and yearly compounding (something that also doesn't exist in nature but isn't particularly important):

Number of periods = ln(current purchasing power/future purchasing power)/ln(1+inflation rate per period)

So, if $1 now buys the same amount as $50 buys later and 4% inflation rate, you get ln(50)/ln(1.04) = about 100 years

The answer depends very much on the inflation rate. At a 2% inflation rate, the answer is about 200 years, and at a 10% inflation rate the answer is about 40 years.

let r be the inflation/interest rate at time t, where t=0 is now, r(0)=1

r: [0,\inf) -> (-1,\inf)

I think r \in L^2 shouldn't limit the practicality

then you just solve

min_t \int_0^t (1+r)*e^t dt = 50

that should do it