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
