Forums > Social Chat > Need math help!
Login/Join to Participatemember
442 posts
Location: Rochester, NY, USA
442 posts
Location: Rochester, NY, USA
Posted:
2^x=y
given y, how do you find x?
My actual problem is:
2 ^ [( X - 16384) / 128] * 4.4 = Y
I know Y. I need to figure out how to reverse the formula and get X. I know I did this sort of thing in high-school, but I don't remember how.
Any math geeks want the challenge?
I'll be using the new formula in a program. So I need to be able to put in any Y and get the resulting X. I just can't remember how to get all the left stuff onto the right side of the equals.
I'd like to end up with an X = blah blah Y.
Thanks in advance.
[ 27. March 2003, 06:45: Message edited by: Pele'sWhippingBoy ]
2^x=y
given y, how do you find x?
My actual problem is:
2 ^ [( X - 16384) / 128] * 4.4 = Y
I know Y. I need to figure out how to reverse the formula and get X. I know I did this sort of thing in high-school, but I don't remember how.
Any math geeks want the challenge?
I'll be using the new formula in a program. So I need to be able to put in any Y and get the resulting X. I just can't remember how to get all the left stuff onto the right side of the equals.
I'd like to end up with an X = blah blah Y.
Thanks in advance.
[ 27. March 2003, 06:45: Message edited by: Pele'sWhippingBoy ]
FYI: I am not Pele. If you wish to reply to me and use a short version of my name, use: PWB.
English? Who needs that? I'm never going to England. - Homer Jay Simpson
SILVER Member
Pooh-Bah
2,014 posts
Location: Galveston, TX, USA
Pooh-Bah
2,014 posts
Location: Galveston, TX, USA
Posted:
x= (log y)/(log2)
it works on the principle that log (2^x)=log y. Log (2^x) = x log 2.
That should get you your answer.
-h
x= (log y)/(log2)
it works on the principle that log (2^x)=log y. Log (2^x) = x log 2.
That should get you your answer.
-h
member
442 posts
Location: Rochester, NY, USA
442 posts
Location: Rochester, NY, USA
Posted:
When I run the formula I use X = 17056 and result in Y = 167.4404
I have tried this formula: (copied from excel)
LOG(2)*(167.4404/4.4)*(128)+16384
And it results in 17850.3151. Not the 17056 that I was expecting.
I know something is messed up. I just can't put my finger on it.
When I run the formula I use X = 17056 and result in Y = 167.4404
I have tried this formula: (copied from excel)
LOG(2)*(167.4404/4.4)*(128)+16384
And it results in 17850.3151. Not the 17056 that I was expecting.
I know something is messed up. I just can't put my finger on it.
FYI: I am not Pele. If you wish to reply to me and use a short version of my name, use: PWB.
English? Who needs that? I'm never going to England. - Homer Jay Simpson
SILVER Member
Pooh-Bah
2,014 posts
Location: Galveston, TX, USA
Pooh-Bah
2,014 posts
Location: Galveston, TX, USA
Posted:
The exact formula would be
X = {128*[(log (y) - log(4.4)]/log(2)} + 16384. I don't know how excel works formula wise, but my guess is you don't have enough parentheses in what you put in. Adding more () never hurts. Part of the problem with your formula is that log2 should be in the denominator, not numerator.
That should help somewhat.
-h
The exact formula would be
X = {128*[(log (y) - log(4.4)]/log(2)} + 16384. I don't know how excel works formula wise, but my guess is you don't have enough parentheses in what you put in. Adding more () never hurts. Part of the problem with your formula is that log2 should be in the denominator, not numerator.
That should help somewhat.
-h
member
442 posts
Location: Rochester, NY, USA
442 posts
Location: Rochester, NY, USA
Posted:
That did it! Thanks.
That did it! Thanks.
FYI: I am not Pele. If you wish to reply to me and use a short version of my name, use: PWB.
English? Who needs that? I'm never going to England. - Homer Jay Simpson
© 1998 - 2024 Home of Poi
All prices converted to United States dollars using XE.com