Got an odd lil question... can I get some help? ~> Suppose you have a scientific calculator where some of the buttons are broken. In fact the only working keys are the digits [0] to [9], [+],[-],[x],[/] and [e^x] and [ln]. Explain how you can use this calculator to calculate the square root of any whole number.
some numbers ln divide some number = e^x
yeah but we dont have ln
Oh crap
this page explains the Babylonian method. It's basically making an educated guess and fine tuning that guess to get more accurate results http://www.deltacollege.edu/dept/basicmath/Babylonian.htm
Ooh nice Jim.
comeonn guyss...
and you call yourselves professors of mathematics
:p
that page has all the answers you need
it was something to do with using indirectly using ln
and it wasn't an estimate :/
on that page, they have the example of sqrt(5) a good guess is 2. Divide 5 by 2 to get 5/2 = 2.5 then notice how 2*2 = 4 is too small and 2.5*2.5 = 6.25 is too big. So the value of sqrt(5) is between 2 and 2.5 average the two: (2+2.5)/2 = 2.25 then you repeat the process all over again with the new guess of 2.25
i "guess" that works :p
well I guess you can say \[\Large \sqrt{x} = x^{1/2}\] but I don't know how to tie it in with \(\Large e^x\) or natural logs. So I'd just stick with the Babylonian method.
how about if you had [ln]?
it's okay guys don't fry your brains :p
still don't know, but I'll think it over
should be easy if you have [ln]
okay thanks anyways <3
oh noes
I believe in you :)
ok if you had both e^x and ln(x) buttons, then you can do this \[\Large \sqrt{x} = e^{\ln(\sqrt{x})} \] \[\Large \sqrt{x} = e^{\ln(x^{1/2})} \] \[\Large \sqrt{x} = e^{\frac{1}{2}\ln(x)} \] ------------------------------------------------------- So for example, \[\Large \sqrt{x} = e^{\frac{1}{2}\ln(x)} \] \[\Large \sqrt{5} = e^{\frac{1}{2}\ln(5)} \] \[\Large \sqrt{5} \approx 2.23606797750042 \]
now wasn't that simple :p
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