Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Log formula
#1
Say I want to work out 10^2.4 (which is 251+loose change). How do I work it out without using the x^y button on my calculator? I'm pretty sure I need logs, but how would I do it?

PS: I need a mathematical formula, series etc.
Reply
#2
ORACLE:
This may be what you're looking for:
Code:
N = A^X
   therefore:
   A = INT(LOG(X) * LOG(N))
   X = INT(LOG(N) / LOG(A))
*****
Reply
#3
Or you could do: (with workings...)
10^2.4
=10^(12/5)
=10^2 · 10^(2/5)
=10^2 · (10^(1/5))^2
=10^2 · (5\/¯10)^2
=(10×10) · ((5\/¯10)×(5\/¯10))

Now you don't need to use the x^y button Tongue
url=http://www.spreadfirefox.com/?q=affiliates&id=60131&t=79][Image: safer.gif][/url]
END OF LINE.
Reply
#4
Not quite moneo... I want to not have to use the log function.

Phydaux's qth root of p^n is better for what I want. Now how do I calculate the 5th root of something in QBasic without using log functions etc? (This is for StatLib, so I have to rewrite every inbuilt function :roll: )
Reply
#5
This uses logs... is that ok?
http://mathforum.org/library/drmath/view/52651.html
am an asshole. Get used to it.
Reply
#6
Quote:Not quite moneo... I want to not have to use the log function.

Phydaux's qth root of p^n is better for what I want. Now how do I calculate the 5th root of something in QBasic without using log functions etc? (This is for StatLib, so I have to rewrite every inbuilt function :roll: )

This calculates integer nth roots: (n\/¯i)
Code:
DIM n AS INTEGER
DIM a AS DOUBLE, b AS DOUBLE

i = 10    'where n is root, i is input
n = 5    '
a = 1
FOR j = 1 TO 10
   b = (1 / n) * ((i / a ^ (n - 1)) + (n - 1) * a)
   a = b
NEXT j
PRINT a
You've still got the exponent there, but integer exponents are simple Smile
Also the more itterations the higher accuracy of depth you get.

HTH

~Phydaux
url=http://www.spreadfirefox.com/?q=affiliates&id=60131&t=79][Image: safer.gif][/url]
END OF LINE.
Reply
#7
You may not want to use i because in the math world, i is the square root of -1
am an asshole. Get used to it.
Reply
#8
I think he wants to know how to calculate 10^2.4 without using ^ on his calculator.

To use logs u must know this formula:
log[(m)^n] = nxlog(m)

So your calculation would be:
Antilog(2.4 x log(10))

Now if u take the log to the base 10 then it would be Antilog(2.4 x 1)...(As log 10 to the base 10 is 1)

Therefore,
Antilog(2.4) = 251.1886432

Ok?
Reply
#9
Quote:You may not want to use i because in the math world, i is the square root of -1
I feel perfectly happy using whatever letters I want.

i is also used for electric current, which is why engineers often use j to mean the square root of -1.

I have also clearly stated what i means so you shouldn't really get confused.

You, may, or may not have noticed that often for a FOR loop, the letter i is used to mean itteration. It's insane that the same symbol can mean different things eh? Tongue
url=http://www.spreadfirefox.com/?q=affiliates&id=60131&t=79][Image: safer.gif][/url]
END OF LINE.
Reply
#10
Quote:So your calculation would be:
2.4 x log(10)

In fact it would give the log of the result. The true result is
exp(2.4Xlog(10))
Antoni
Reply


Forum Jump:


Users browsing this thread: 2 Guest(s)