how would you solve 5^x=x ?
824446
Not sure... but the answer is a complex number, I'll tell you that. There is no real solution for 5^x = x. In the equation a^x=x, a has to be less than e and greater than 0 for there to be a real solution.
894234
a^(m/n)=(a^m)^1/n\nor (a^1/n)m\nsay 8^(5/3)\n=(8^1/3)^5\n=2^5=32
1136055
Here's how we can solve the problem logically, without doing all the multiplication.\n\nFirst of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.\n\nLet's create two new temporary variables: let a = x + 5y, and b = 3z + w.  That reduces the problem, for the moment, to (a + b)^6.\n\nNow, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.\n\nSo in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6.  Note that, in each term, the sum of the powers is 6.\n\nNow, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.\n\na^6 becomes (x+5y)^6, which will have 7 terms.\na^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.\na^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.\na^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.\na^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.\nab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.\nAnd, finally, b^6 will have 7 terms.\n\n7+12+15+16+15+12+7 = 84.\n\nHope that made sense.
141310
to solve for X, you have to separate X from everything else. in that problem 2X+3X=5, you would combine the 2X and 3X to get 5X. so the equation end up as 5X=5. from there, divide both side by 5. 5X/5=X and 5/5=1 so X=1.
492387
This has no solution, 2^x would be equal to a negative number. That can't happen.
528657
x^4 - 8x^2 + 15 = 0\n(x^2 - 5)(x^2 - 3) = 0\nSo x^2 = 5 or x^2 = 3.\nSo x = sqrt(5) or x = -sqrt(5) or x = sqrt(3) or x = -sqrt(3).\n\nTo purdue_engineer:\nIf it makes you feel any better, I've been screwed a bit lately.  One person copied my answer and he ended up getting best answer.  Another person was wrong but bashed my answer and so she got best answer instead of me!\n\nBut anyway, sorry!  I'm a teacher, and I just can't help myself.
283872
x^x^x^x..... like that upto infinity right\nok then we will take one x out from the  equation and the rest of them will be assigned the value 9\n\ni.e x^9=9\n\nx=9^1/9\n\nor x=1.27\n\nhope u understood
707418
20x^2+60x + 45\n= 5(4x^2+12x + 9) \n= 5((2x)^2+2(2x)3+3^2) = 5(2x+3)^2\n\n3x^2+12x+12 = 0\n=> x^2+4x+4 = 0 (divide by 4)\n=>(x+2) ^2 = 0 or x = -2, -2 (double root)\n\n7x^2-28 =0\n= x^2 = 4\n=> x = 2 or -2\n\narea = xy = 5*2 = 10cm^2
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2
198477
x^2-6x+5=\n\nx^2-5x-x+5=  [presenting -6x as -5x-x]\n\nx(x-5)-1(x-5)= [getting out common factor]\n\n(x-1)(x-5)\n\nTo see that it was done correct put any small number in the original expression and the result and compare them. If you did right, you should get two numbers which are equal to each other.\n\nPutting x=1 on the left side(the original expression):\n\n1^2-6*1+5=1-6+5=-5+5= 0.\n\nPutting x=1 on the right side(the result):\n\n(1-1)(1-5)=0*(-4) = 0.\n\nWe got on both sides the number 0, as expected, which probably means we did it right.\nTher final answer is:\n\n(x-1)(x-5)\n\nHope that helps.
256804
5x^2  -10 x +5\ncan be factorised into (x-1)(5x-5)\nsimplified to 5(x-1)(x-1)\n\nthus square root of given sum becomes square root of \n5(x-1)^2\n\nwhich is (x-1)*5^1/2
978189
e^(ln5)=x   .  Now express this in log form by thinking of ln(5)\n                       as just an exponent.\n\n   lnx=ln5   so obviously\n\n   x=5.
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
471375
First, there is a difference between\n1) 3^x-5=27 and\n2) 3^(x-5)=27    1) means raise 3 to the x power, then\n\nsubtract 5 , 2) means subtract 5 from x and raise 3 to that power.\nCheck your original and be sure. In math, we're extremely\nand totally precise. It's part of the whole academic discipline. \n  About you log key : Calculators are different so you will have to experiment a little. Key in 10 then hit log. You\nmust get 1. For now ignore confusing issues with the use\nof ln, you don't need to know that. To make sure, key in\n100 and hit log. You must get 2.\n  No for anti log or inverse log experiment some more and/or get help. Key in 1 and punch your inverse log\nsequence or key. The answer must be 10. Again do it for 2.\nThe answer must be 100. When you have mastered that then\nproceed to the problem.\n Assume your problem is 1). Then add 5 to both sides of the equation giving you 3^x=32. Now solve\n x(log(3))=log(32)\nx=log(32)/log(3) \nTake the log of 32 giving 1.505  .\nTake the log of 3 giving .477  so\nx=1.505/.477 = 3.155\n\n If the problem was actually 2) then we have\n(x-5)(log(3))=log(27)\n(x-5)=log(27)/log(3) = 1.43/.477 = 3.00\nx=8
879323
x^4+4x^3+4x^2-4x-5 \n= (x^4 - x^2) + (4x^3 - 4x) + (5x^2 - 5)\n= x^2(x^2 -1) + 4x(x^2 -1) + 5(x^2 -1)\n= (x^2 -1)(x^2 + 4x + 5)\n= (x+1)(x-1)(x^2 + 4x + 5)\nSo, -1,1 and the roots of x^2 + 4x + 5 which are:\n\n-4/2 ± √(16 - 20) /2 = -2  ± √(-4) /2 = -2 ± 2i/2 = -2 ± i\n\nSo,\n1, -1, -2+i, -2-i
912352
I believe the answer should be \n\nlog{ [ 5 +/- sqrt(39)i ]/8} / log 2.\n\nSteps are as follows:\n\n2^x + 1/2^x = 5/4\n(2^x * 2^x + 1)/2^x = 5/4\n4(2^x * 2^x + 1) = 5 * 2^x\n4 * 2^x * 2^x + 4 = 5 * 2^x\n4(2^x)^2 - 5(2^x) + 4 = 0\n\nlet 2^x be y\n\n4y^2 - 5y + 4 = 0\n\nsolve it by the equation, you get\n\ny = [5 +/- sqrt(25 - 4*4*4) ]/2*4\ny = [5 +/- sqrt(-39) ]/8\n\nsince y = 2^x\n\n2^x = [5 +/- sqrt(-39) ]/8\nxlog(2) = log {[5 +/- sqrt(-39) ]/8}\nx = log {[5 +/- sqrt(-39) ]/8} / log (2)\nor\nx = log {[5 +/- sqrt(39)i ]/8} / log (2)
97164
5 x squared would just be 5x^2\n\nx^-2  =  1/x^2\n\nSo, y negative 6 exponent  =  1/y^6