Is it possible to solve for X?
April 25, 2012 2:45 PM Subscribe
(Y)(S^N) = (M)(X^(N+E)) - (Y)(X^N) Is it possible to solve for X?
I'm working on a manuscript (psychophysics), and it would be nice if I could solve for X in the following equation:
(Y)(S^N) = (M)(X^(N+E)) - (Y)(X^N)
the damn E term seems to be ruining things. Without it, it's simple enough.
My math isn't the greatest, so I could really use some help!
thanks!
I'm working on a manuscript (psychophysics), and it would be nice if I could solve for X in the following equation:
(Y)(S^N) = (M)(X^(N+E)) - (Y)(X^N)
the damn E term seems to be ruining things. Without it, it's simple enough.
My math isn't the greatest, so I could really use some help!
thanks!
Response by poster: What if I set E to 0.5 (while leaving N as a variable), would it be possible then?
posted by spacediver at 3:26 PM on April 25, 2012
posted by spacediver at 3:26 PM on April 25, 2012
Response by poster: actually, I just realized I have values for N, E, S, and M. But I sense, based on my doodlings, that it's still impossible.
posted by spacediver at 3:32 PM on April 25, 2012
posted by spacediver at 3:32 PM on April 25, 2012
Are the constants (specifically N,E) whole numbers? If so, does N+E<5>
(Also, do you need an equation or to solve for specific values? Could you try solving for Y in terms of X and then graphing it?)
posted by oracle bone at 3:41 PM on April 25, 2012
(Also, do you need an equation or to solve for specific values? Could you try solving for Y in terms of X and then graphing it?)
posted by oracle bone at 3:41 PM on April 25, 2012
Actually, I think you can. Use natural logs on both sides then I think you can work the algebra so lnX is all by itself.
I'm supposed to be finishing my term grades so I can't really finish it up, but I think it works from there.
posted by dness2 at 3:41 PM on April 25, 2012
I'm supposed to be finishing my term grades so I can't really finish it up, but I think it works from there.
posted by dness2 at 3:41 PM on April 25, 2012
Sorry, that was supposed to say "N+E<5?"
posted by oracle bone at 3:42 PM on April 25, 2012
posted by oracle bone at 3:42 PM on April 25, 2012
Do you want a numerical solution? That should be possible if you have all the other values. What you can't get (in general) is an analytic solution in closed form.
posted by Obscure Reference at 3:47 PM on April 25, 2012
posted by Obscure Reference at 3:47 PM on April 25, 2012
Response by poster: oracle: the constants are not integers, but they're all positive values. N + E does add up to less than 5.
I've already graphed Y in terms of X, so I could do it crudely to look up specific values, but would be ideal to be able to graph X in terms of Y based on a continuous function.
dness: thanks for the suggestion, but I barely understand the natural logarithm
posted by spacediver at 3:47 PM on April 25, 2012
I've already graphed Y in terms of X, so I could do it crudely to look up specific values, but would be ideal to be able to graph X in terms of Y based on a continuous function.
dness: thanks for the suggestion, but I barely understand the natural logarithm
posted by spacediver at 3:47 PM on April 25, 2012
Response by poster: obscure: not quite sure what you mean. What i need to be able to do is to plot X as a function of [(Y+1) - (Y)]
posted by spacediver at 3:49 PM on April 25, 2012
posted by spacediver at 3:49 PM on April 25, 2012
Are you sure? Because (Y+1)-(Y)=1. Plotting X as a function of 1 doesn't make sense.
posted by zeptoweasel at 4:07 PM on April 25, 2012
posted by zeptoweasel at 4:07 PM on April 25, 2012
What they mean by natural log is this.
Y^X = A
ln(Y^X) = ln(A)
X*ln(Y) = ln (A)
X = [ln(A)] รท [ln(Y)]
The trick here is that ln(Y^X) is the same as X*ln(Y).
posted by spikeleemajortomdickandharryconnickjrmints at 4:28 PM on April 25, 2012
Y^X = A
ln(Y^X) = ln(A)
X*ln(Y) = ln (A)
X = [ln(A)] รท [ln(Y)]
The trick here is that ln(Y^X) is the same as X*ln(Y).
posted by spikeleemajortomdickandharryconnickjrmints at 4:28 PM on April 25, 2012
Oh, and this works for any log base, not just natural log.
So log(Y^X) = X*log(Y).
posted by spikeleemajortomdickandharryconnickjrmints at 4:29 PM on April 25, 2012
So log(Y^X) = X*log(Y).
posted by spikeleemajortomdickandharryconnickjrmints at 4:29 PM on April 25, 2012
And just looking at it, 1 minute before class starts, it seems to me that you can solve for X. I'll break out a pen and paper when I'm out of class and let you know if I solve it.
posted by spikeleemajortomdickandharryconnickjrmints at 4:30 PM on April 25, 2012
posted by spikeleemajortomdickandharryconnickjrmints at 4:30 PM on April 25, 2012
Nope, I can't do it. Might be possible, but I'm not seeing it.
posted by spikeleemajortomdickandharryconnickjrmints at 4:33 PM on April 25, 2012
posted by spikeleemajortomdickandharryconnickjrmints at 4:33 PM on April 25, 2012
Response by poster: zepto: Y refers to a function there not a variable.
spikelee: i see what u mean
posted by spacediver at 4:36 PM on April 25, 2012
spikelee: i see what u mean
posted by spacediver at 4:36 PM on April 25, 2012
Response by poster: zepto: to clarify:
I need to solve for f(Y+1) - f(Y)
posted by spacediver at 4:41 PM on April 25, 2012
I need to solve for f(Y+1) - f(Y)
posted by spacediver at 4:41 PM on April 25, 2012
Response by poster: thanks for all the help here, I really appreciate it. I'm gonna try the crude approximation method - should be good enough :)
posted by spacediver at 5:03 PM on April 25, 2012
posted by spacediver at 5:03 PM on April 25, 2012
I am not a math person. Is there any reason you can't plug in numbers and simplify?
(Y)(S^N) = (M)(X^(N+E)) - (Y)(X^N)
If:
y=5
s=3
n=2
m=10
e=1
so:
(5)(3^2) = (10)(X^(2+1)) - (5)(X^2)
=(5)(3^2) = (10)(x^3)-(5)(x^2)
= 5(9) = 10x^3 - 5x^2
= 45 = 10-5 (x^3)(x^2)
= 45 = 5x^5
= 9 = x^5
= 9^.20 = x
=1.55 = x
posted by triggerfinger at 5:39 PM on April 25, 2012
(Y)(S^N) = (M)(X^(N+E)) - (Y)(X^N)
If:
y=5
s=3
n=2
m=10
e=1
so:
(5)(3^2) = (10)(X^(2+1)) - (5)(X^2)
=(5)(3^2) = (10)(x^3)-(5)(x^2)
= 5(9) = 10x^3 - 5x^2
= 45 = 10-5 (x^3)(x^2)
= 45 = 5x^5
= 9 = x^5
= 9^.20 = x
=1.55 = x
posted by triggerfinger at 5:39 PM on April 25, 2012
Response by poster: trigger: that would work if i were looking for the value of x at a particular y, but in my case, i need to find the value of x at any given y.
Anyway, I've figured out the crude approximation solution and it works beautifully.
This helped.
posted by spacediver at 5:56 PM on April 25, 2012
Anyway, I've figured out the crude approximation solution and it works beautifully.
This helped.
posted by spacediver at 5:56 PM on April 25, 2012
I'm back. If I did it right the answer is:
X=Y/M-[e^(N+E) /S^(N^2)]
Like Spikeetc. Was saying, take the ln of both sides, get lnX by itself then raise both sides to e.
posted by dness2 at 5:56 PM on April 25, 2012
X=Y/M-[e^(N+E) /S^(N^2)]
Like Spikeetc. Was saying, take the ln of both sides, get lnX by itself then raise both sides to e.
posted by dness2 at 5:56 PM on April 25, 2012
Wolfram Alpha is always useful if you've forgotten your math . . .
(I changed E to F because W|A was interpreting E as the math constant e.)
You can rather easily solve for Y--that might be useful.
posted by flug at 6:31 PM on April 25, 2012 [1 favorite]
(I changed E to F because W|A was interpreting E as the math constant e.)
You can rather easily solve for Y--that might be useful.
posted by flug at 6:31 PM on April 25, 2012 [1 favorite]
Response by poster: dness: I don't think that's correct. When I plot X in terms of Y and compare Y in terms of X, they don't correspond.
flug: that's a very nifty tool. I already have the solution for X in terms of Y though. The formulation in the original post was based on that original formulation - i had already done some re-arranging to make it more digestible.
thank you both for the efforts though :)
posted by spacediver at 9:28 PM on April 25, 2012
flug: that's a very nifty tool. I already have the solution for X in terms of Y though. The formulation in the original post was based on that original formulation - i had already done some re-arranging to make it more digestible.
thank you both for the efforts though :)
posted by spacediver at 9:28 PM on April 25, 2012
Response by poster: dness, in case you were wondering, I checked both
X=[Y/M]-[e^(N+E) /S^(N^2)]
and
X=Y/[M-[e^(N+E) /S^(N^2)]]
posted by spacediver at 9:29 PM on April 25, 2012
X=[Y/M]-[e^(N+E) /S^(N^2)]
and
X=Y/[M-[e^(N+E) /S^(N^2)]]
posted by spacediver at 9:29 PM on April 25, 2012
This thread is closed to new comments.
posted by vacapinta at 2:56 PM on April 25, 2012