OpenEuphoria

2. Re: Rather large number - oops

• Posted by Rolf Schroeder <schroeder at DESY.DE> Nov 25, 1998
• 425 views

Molasses wrote:
>
> Thanks to those that answered,
> but I think the equation should have read :
>
> (16777216^480000) / 756864000
>
> 16m to the power of 480000, instead of the other way round, sorry.
> Of course (not that I understood any of the math of it), as Hawke
> said, 3.63e14 would be a better divisor.
>
> The equation would then be:
> 256^3^480000/3.63e14
>
> is this right?
> and could someone answer it please? 
>
> TIA
>
> -molasses

OK, again: (look at the end!)
--
--
--  d = (16777216)^480000 / 3.63e14
--
--  using log: log(d) = 480000 * log(16777216) - log(3.63e14)
--  dec. log: dlog(d) = log(d)/log(10)
--
    atom  a,  b,  c
    atom ln, ld, dec_exponent, dlog_mantissa, mantissa

    a = 480000
    b = 16777216
    c = 756864000

    ln = b * log(a) - log(c)
    ld = ln/log(10)
    printf(1,"Dec. log     : %25.15f\n", ld)
    dec_exponent = floor(ld)
    printf(1,"Dec.exp.     : %25.15f\n", dec_exponent)
    dlog_mantissa = ld - dec_exponent
    printf(1,"DLog.Mantissa: %25.15f\n", dlog_mantissa)
    mantissa = power(10,dlog_mantissa)
    printf(1,"Mantissa     : %25.15f\n", mantissa)
--
--  The result of your number d is:
--
    puts(1,"The result:\n\n")
    printf(1,"(16777216)^480000 / 3.63e14 = %10.5f*10^%d\n",
                {mantissa,dec_exponent})
--
--
--  This will be the result:
--
--  (16777216)^480000 / 3.63e14 =    3.22507*10^95315402
--
Have a nice day, Rolf