if y = e^x-1 then x =???
493962
depends\n\nif that is\ny = e^(x-1)\n\nor\n\ny = e^x - 1 (remeber no superscript!)\n\nthey are\n\nx = ln y + 1\n\nor\n\nx = ln (y + 1)
1098401
Check the problem again.  It doesn't simplify much ... I would have expected the terms with e^-x to either both be subtracted or added to one, not having the diferent signs you show.\n\nIt's not zero even how you have it written.  If x=0, the result is (1+1)(1+1)- (1+1)(1-1)/(1+1)^2.  Or 4 - 0 = 4.  The answer isn't 4 either (for all x).
42255
The web page [1] below gives the solution to\n\n    x^x = 1333\n\nas\n\n    x = e^(W(ln(1333)) = 4.669202878...\n\nwhere W is Lambert's W function (see web page [2] or more generally web page [3]).\n\nSo for x^x = y if y is in the appropriate range then x would be e^(W(ln(y))).
749046
well the term you've quoted is a constant...\n\nso (e^2/1+e^2)x would be the answer to that question.\n\nbut i'm guessing you typed it wrong, or are confused.
809369
if y' = x/12\nthen\ny= x^2 /24\ns
709175
If you need a positive result, then y will only be a real number when x - 1 is >= 0.\n\nx - 1 >= 0\nx >= 1\n\nSo x must be greater or equal to 1.
775564
if ln(x)=y, then by the definition of ln, x=e^y\ne^(ln(x)) for ln(x)=y is e^y\nso e^ln(x)=e^y which we already showed =x which is what we wanted to show. Hope this helps. If not, email me.
602488
1-x = -(x-1)\nx^3 - 1 = (x-1)(x^2+x+1)\nAnswer: -(x-1)(x^2+x+1)
13683
Let e^x equal some variable (let's use y).\ny+1/y=4\nMultiply both sides by y.\ny^2+1=4y\ny^2-4y+1=0\nNow you can use the quadratic formula to find y.\ny=2+2√2 and 2-2√2\nBut don't forget that we said y=e^x, so make sure to solve for x.\nln(2+2√2)=x (In this case, y cannot be negative because one cannot take the natural log of a negative number)\nx≈1.175
973577
given x < y\n\nnot sure if this would be using mathmatical induction but this is how i would do it\n\ntake\nx < 1/2 (x+y) < y\n2x< x + y < 2y\nx < y < 2y\n\nx<y is stated to be true and 2y is certainly bigger than y\ntherefore x < 1/2 (x+y) < y
88953
=\na(x-y)+x(x-y)\n=(a+x)(x-y)\ndone
155074
I get the same y=-92/79, x=127/79\n\nI thought this couldn't be right.
869813
(y--1)/(-2--1)=(x-1)/(2-1)\n=(y+1)/(-2+1)=(x-1)/(1)\n=>y+1=-1(x-1)\ny+1=-x+1\ny=-x is the required equation
900593
y=f(x)= a*x*e^(b*(x^2)),\n y’=a*[e^(b*(x^2))+x*b*2*x*e^(b*(x^2))],\n y’=a*[1+2*b*x^2]*e^(b*(x^2))=0,\nThe value of e^(b*(x^2)) is always >0, also ‘a’ is not =0. \nthen 1+2*b*x^2=0, while x=3, b=-1/(2*3*3)=-1/18\n& 4=a*3*e^(-3*3/18), hence a=(4/3)*sqrt(e)\nis it really MAX?\ny’(0)=a>0 means increasing function, thus YES!
1053953
Divide both sides by 11.\ne^x = 11\nThen take the natural log (ln) of both sides.  Any good scientific calculator can do this.\nln(e^x) = ln(11)\nx = ln(11)
721733
if y is always increasing then y' is always >=0 (1) \ny'=3ax^2+2bx+c \ndelta': D=b^2-3ac (2)\nwe also have a>0 (3)\n(1),(2)&(3)--> D <=0-->b^2-3ac <=0-->b^2<=3ac (qed)\nNote: we have b^2 <=3ac, not b^2 < 3ac,for example : function y=(x-1)^3
866301
because y- x > 0\n\n|y-x | = y- x (positive value)\n\nx-y < 0\n\nso | x-y| = y-x (as absolute value positive)