your problem is somewhat insolvable, since the square root of 5 has
an infinite number of digits, so it cannot be represented correctly
on any computer (in decimal or binary notation).
For arbitrary precision calculations with Fixnum exponents, there's
bigdecimal.

require "bigdecimal"

def calc(n,prec)
# prec is the precision of the sqrt calculations
res=(BigDecimal.new("2") ** n)*BigDecimal("5").sqrt(prec)
end

puts calc(10000) => 0.44610[several lines of digits]*10^3011 (for prec=10)

However, don't believe in too many of these digits...
If you still want to multiply by square roots accurately, it might be a good
idea to look at continued fractions - every square root has a continued
fraction representation that eventually ends in a periodic pattern.
There is an introduction to arithmetic with them at
_http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html_
(http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html)

your problem is somewhat insolvable, since the square root of 5 has
an infinite number of digits, so it cannot be represented correctly
on any computer (in decimal or binary notation).
For arbitrary precision calculations with Fixnum exponents, there's
bigdecimal.

require "bigdecimal"

def calc(n,prec)
# prec is the precision of the sqrt calculations
res=(BigDecimal.new("2") ** n)*BigDecimal("5").sqrt(prec)
end

puts calc(10000) => 0.44610[several lines of digits]*10^3011 (for prec=10)

However, don't believe in too many of these digits...
If you still want to multiply by square roots accurately, it might be a good
idea to look at continued fractions - every square root has a continued
fraction representation that eventually ends in a periodic pattern.
There is an introduction to arithmetic with them at
_http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html_
(http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html)