# [SciPy-user] Can SciPy compute ln(640320**3 + 744)/163**.5 to 30 places?

Dick Moores rdm at rcblue.com
Sun Jan 21 23:34:09 CST 2007

```At 09:22 PM 1/21/2007, you wrote:
>On 1/21/07, Dick Moores <rdm at rcblue.com> wrote:
>
> > OK, I'll reclarify. Of course, I'm trying to learn clnum. I thought
> > that was understood. Please show me how to use clnum to compute
> > (5/23)**(2/7) (that's 5/23  to the  2/7 power), with a precision of
> > 50. Surely the answer is not
> > mpf('0.06211180124223602484472049689440993788819875776397515527949',55)??
> > Just by inspection this is clear, is it not? A number between 0 and 1
> > raised to a power between 0 and 1 will get closer to 1, right? (5/23
> > is 0.217391304348) I believe you've given me (5/23.)*(2/7.).
>
>Sorry, my bad: one misplaced parenthesis and me being terribly sloppy
>in not actually checking the numerical value (I just hastily saw
>0..6.. and didn't look further).
>
>In [2]: import clnum as n
>
>In [3]: n.set_default_precision(50)
>
>
>This is what I repeated to you like a mindless idiot:
>
>In [4]: n.exp(n.log(n.mpq('5/23')*n.mpq('2/7')))
>Out[4]:
>mpf('0.06211180124223602484472049689440993788819875776397515527949',55)
>
>
>This is the correct form (note position of parenthesis):
>
>In [5]: n.exp(n.log(n.mpq('5/23'))*n.mpq('2/7'))
>Out[5]: mpf('0.6466073240654112346263901524238077888294103593272266200345',55)
>
>
>Again, my apology for the confusion, it was entirely my fault.

No, no. I kept screwing up my question.

You've been of immense assistance. Thank you.

Dick

```