Yes, I do think that's a good solution. On Thu, May 14, 2015 at 2:32 PM, harven <[email protected]> wrote:
> Thanks for the answer. I understand now. So log(2,x) is log(x)/log(2) and > the denominator is a Float64 so we don't get the desired precision. So I > can just write > > julia> with_bigfloat_precision(10_000) do > log(big(2)^10_000)/log(big(2)) > end > 1e+04 with 10000 bits of precision > > A solution maybe is to promote the base to the same type as the argument? > > julia> mylog(b,x) = log(x)/log(oftype(x,b)) > mylog (generic function with 1 method) > > julia> with_bigfloat_precision(10_000) do > mylog(2,big(2)^10_000) > end > 1e+04 with 10000 bits of precision >
