how to solve X ln X = 4, is there any closed form solution? thanks?
362561
the real value of x = 2 becuause x times x equals 4 so 2x 2 equals 4
817874
Remember that log x - log y =log (x/y) to rewrite that equation. Then, using base 10, you have the equation 10=(x+4/ x+3). Solve.
760030
okay well pretty much you have to solve for  x\nso inorder to get X by itslef you have to -4 from -8 like so:\n2x + 4-4 = –8-4\n2x=-12\nnow you still have 2X but now you can divide the 2 into -12\nx= \nI won't give you the answer but you should be able to find the answer!!!
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
240063
1. Solve for the positive value of x: x/3 - 4/x = 4/3\nx/3 - 4/x = 4/3\n3x((x/3) - (4/x) = (4/3))\nx^2 - 12 = 4x\nx^2 - 4x - 12 = 0\nx = [-(-4) +- sqrt((-4)^2 - 4*1*(-12))]/2\nx = [4 +- sqrt(16 + 48)]/2\nx = [4 +- 8]/2\nx = -2 or 6 (but 6 is the only positive value)\n\n2. Find the coordinates of P', the image of P(3,-1) under the transformation (x,y) --> (-y,-x)\nMy guess is that P' is (1, -3).\n\n3. Which equation has both 3 and 6 as roots?\nYour expression will come to (x - 3)(x - 6) = 0, or x^2 - 9x + 18 = 0, so given:\nA) *(square root of) x-2* = x-4\nB) *x-2* = 4-x\nC) *x-2* = 3/x\nD) *x-2* = x/3\n\nCheck A: *(square root of) x-2* = x-4\nsqrt(x - 2) = x - 4\nx - 2 = (x - 4)^2\nx - 2 = x^2 - 8x + 16\nx^2 - 9x + 18 = 0\n\nCheck B *x-2* = 4-x\nx - 2 = (4 - x)^2\nx - 2 = 16 - 8x + x^2\nx^2 - 9x + 18 = 0\n\nCheck C: *x-2* = 3/x\nx - 2 = 9/(x^2)\nx^3 - 2x^2 = 9\nIt doesn't simplify the same way, so you'll need to "plug and chug" with 3 and 6:\n3^3 - 2*(3^2) ?= 9\n27 - 2*9 ?= 9\n27 - 18 ?= 9\n9 = 9\n-----\n6^3 - 2*(6^2) ?= 9\n216 - 2*36 <> 9\n("C" doesn't satisfy the conditions.)\n\nCheck D: *x-2* = x/3\nx - 2 = (x^2)/9\n9x - 18 = x^2\nx^2 - 9x + 18 = 0\n\nSo A, B, and D work; C doesn't.\n\n4. Looks good to me!
805634
It is easier (by far) to do this with logarithmic differentiation:\ndy/dx=y*d/dx[ln y].\n\nHere, with y equal to your expression,\nln y=ln 10 +ln x +4ln(x-1)\n-3ln(x-2)-2ln(x+1),\nso the derivative of the logarithm is\n1/x +4/(x-1) -3/(x-2)-2/(x+1).\nThe first derivative is now your original expression times this.\n\nTo find the second derivative, use the product rule. The derivative of y is already known, and the second derivative of the logarithm is easy.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
914071
dy/dx +4y = 6\n\nSo dy/dx = 6 - 4y\n\nso dx/dy = 1/(6 - 4y)\n\nThus x = -¼ ln(6 - 4y) + c\nSo ln(6 - 4y) = -4x + k Where k = 4c\nThus 6 - 4y = e^(-4x + k) = Ae^-4x where A = e^k\nSo 4y = 6 - Ae^(-4x) \nie y = 1.5 - Be^(-4x) (where B = A/4)\n\nCheck dy/dx = 4Be^(-4x) = 4 (1.5 - y) = 6 - 4y\nNow y(0) = 0\nSo 0 = 1.5 - Be^0\nB = 1.5\n\nSo the solution being sought is:\n\ny = 1.5(1 - e^(-4x))
19421
X^5 = 98\n=> ln(X^5) = ln98\n=> 5lnX = ln98\n=> lnX = (ln98)/5\n=> X = e^((ln98)/5)\nX = 2.50175752714084352...
350639
7^(2x + 1) = 3^(x + 2)\nlog(7)(x + 2) = log(3)(2x + 1)\nIn the second step, you switched the x+2 and 2x+1\nLine 2 should be\nlog(7)(2x + 1) = log(3)(x + 2)\nContinue solving as you did before and you will get the correct answer.\n\nAlso, you wouldn't write ln9 just because it is not in simplest form\nln(9) = ln(3^2) = 2ln(3)\nSimilarly for ln(49) = 2ln(7)
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.
634689
here \n ln2+ln(x-6)=ln(x+2)\n \n ln(2*(x-6))=ln(x+2)\n\n i.e 2x-12=x+2\n   \n     x=14
880448
ln(x+6)-ln5=ln(x+6)/5  difference rule\nln(x+6)/5=ln(2x-7)  given\nsince the logs are equal removing the log sign\n(x+6)/5=2x-7\nmultiplying by 5\nx=6=5(2x-7)\nx+6=10x-35\nadding -6\nx=10x-41\nadding-x\n0=9x-41\nadding 41\n9x=41\ndividing by 9\nx=41/9
229078
Its curious that nobody got it right above.\n\nln(x) + ln(y) = 1 + y'\n1/x + y'/y = 1 + y'\n1/x - 1 = y' (1 - y)\n\nand so on
19796
3^2x - 5(3^(x+4)) = 0\n=> 3^2x = 5(3^(x+4))\n=> Ln(3^2x) = Ln(5(3^(x+4)))\n=> Ln(3^2x) = Ln5 + Ln(3^(x+4))\n=> 2xLn3 = Ln5 +(x+4)Ln3\n=> 2xLn3 = Ln5 +xLn3 +4Ln3\n=> xLn3 = Ln5 +4Ln3\n=> x = Ln5/Ln3 +4Ln3/Ln3\n=> x = Ln2 + 4
728396
1.\nln(5x + 2) + ln x = ln e\nln(5x + 2) + ln x = 1\ne ^ (ln(5x + 2) + ln x) = e ^ (1)\n(e ^ ln(5x + 2))(e ^ ln x) = e\n(5x + 2)(x) = e\n5x^2 + 2x - e = 0\nx = (-2 +/- sqrt(4 - (4)(5)(-e))) / 2(5)\nx = (-2 +/- sqrt(4 + 20e)) / 10\nx = (-2 +/- (2)sqrt(1 + 5e)) / 10\nx = (-1 +/- sqrt(1 + 5e)) / 5\ntest answers to ensure ln of positive number\nx = (-1 + sqrt(1 + 5e)) / 5\n\n2. \nln(x + 1) / ln(x - 1) = 2\nln(x + 1) = (2)ln(x - 1)\nln(x + 1) = ln((x - 1)^2)\nx + 1 = (x - 1) ^ 2\nx + 1 = x^2 - 2x + 1\n0 = x^2 - 3x\nx(x - 3) = 0\nx = 0 or x = 3\ntest answers to ensure ln of positive number\nx = 3
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4