How to solve x^x=y. Simple explanation, please.?
830737
I do not understand what it is you are trying to solve. Are yoiu trying to find the inverse fuction? If you are that will be extremely difficult. If you seek the derivative then it has been doone above for you.\n\nIf you seek the integral then it would probably requiire a power series But as it is your problem is not a problem as you have not stated what you want solved.\n\ny = x^x\n= e^xlnx\n\ndy/dx = (lnx + 1)x*x\n= (lnx +1)*y\nSo lnx = 1/y *dy/dx - 1\nie x = 1/e * e^(1/lny * dy/dx)\nWhence x = 1/e * e^ [1/(lny* dx/dy)]\n\nNow I find that unconvincing as to get dx/dy I need to know x in terms of y Its looped onto itself!!!! In fact it is a differential equation not a function. ( x = 1/e * e^ [1/(x' . lny)]
223613
dont the value of x have to be the same? to the person above me..\n\ni thought that x^2 means x to the power of 2 (x squared)\nooo wait i dont know im confusing myself. im only 13 and not that smart but these are my answers now.\n\n\nfirst i got x= negetive eight.\nthen i got x= 2 and -8\n\n\nand gary, wow thats befuddling.\nyour a nerd.\n\nwhat are you kidding me?? thats so not seventh grade math! thats algebra, and im only in pre algrebra.
685420
x-y = 4\nx = 4 + y\nx-4 = y\n\nSo all pairs of the for (x, x-4).
444064
So, for a) & b) the two things that you should notice are:\n1 - The exponents inside the second polynomial alternate, meaning that as the powers of the x variable decrease, the powers of the y variable increase, both by increments of one.\n2 - The answer will only contain the variables x and y raised to the same power, which is one more than the leading exponent on the x variable.  \n\nSince (x + y)(x^2 - xy + y^2) = x^3 + y^3\n and  (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = x^5 + y^5,\nthen for any given polynomial the pattern is written out...\n(x + y)(y^2n - x^(2n-1)y + x^(2n-2)y^2 + ... + x^2y^(2n-2) - xy^(2n-1) + y^2n = x^(2n+1) + y^(2n+1)
955331
3^-x=2\n-x ln3=ln2\n-x=ln2/ln3\n2x=-2ln2/ln3=-1.2618595071429148741990542286855\n3^-1.2618595071429148741990542286855=.25
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
492313
x^2*y"-x*y' +4*x*e^(x^3)=0\n\nx[x*y"-y' +4*e^(x^3)]=0\n\n=> x=0 or x*y"-y' +4*e^(x^3)=0\n\nx*y"-y' +4*e^(x^3)=0\n\n=>\n\n∫[x*y"-y' +4*e^(x^3)].dx=0\n\n=> [-4*x*Gamma[1/3, -(x^3)]]/[3(-(x^3))^(1/3)]=0\n\nwell even my computer couldn't solve that, so let the answer be x=0.
961234
Let x = y^(1/3)\nx^2 - 9x + 8 = 0\n(x - 8)(x - 1) = 0\nx = 8 or 1\ny^(1/3) = 8 or 1\ny = 2 or 1
984409
The goal here is to find the equation with x on one side and everything else on the other.  It's like the equation you have, but instead of y= you want x=.  So you need to "solve for x."  Think of y as just a number, and you need to find what x =.\n\nStep by step:\n\ny=4x+4\n\nSubtract 4 from each side:\n\ny-4=4x\n\nDivide both sides by 4\n\n(y-4)/4=x\n\nx=(y-4)/4\n\nThere's your equation!  Hope this helps.
469518
There is an obvious 1-dimensional continuum of solutions to this equation: the line y=x.\nHowever, there are a continuum of more solutions, most of which are nontrivial:\nNote that log(x^y) = y*log(x). Thus log(x)/log(y) = x/y. This implies that y = x^(y/x) and similarly x = y^(x/y). This is the general form of the solution where the obvious solution is the restriction x/y = 1. \nIf we let x/y = 2, we get 2y = y^2, or y(y - 2) = 0 which gives nondegenerate solution point (4, 2) as mentioned previously. Setting x/y = 3 gives us 3y = y^3 or y^2 - 3 = 0, which gives the nontrivial solution point (3*Sqrt[3], Sqrt[3]) and so on.
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
160067
When you substitute with y=x^(1/6), you're assuming :\nI) x>0 (which should be put as a starting condition)\nII) y>0 . This is to be considered after solving the 2nd degree equation, as you get a fake solution y= -3 that you must reject as non-acceptable.\n\nThe only acceptable solution is x= 1/64
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
612878
xy = 4 then y = x/4\n\nx^2 +  4y^2 = 20\nx^2 + 4 x^2/16 = 20\nx^2 + x^2/4 = 20\n4x^2 + x^2 = 80\n5x^2 = 80\nx^2 = 80/5\nx^2 = 16\n\ntherefor\nx = 4\nthen \ny=1
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
810076
11^(0.35+7) = 3 this is a contradiction\n\nx=3 is a solution but the rest is just a mess.\n\nDo you have any idea of what you are doing?