solve for x: 4(b)^x = a^x?
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
363348
Are we solving for x?
171139
whenever you divide numbers that have exponents, this rule will apply:\n\n(x^a)/(x^b) = x^(a-b)\n\nso in another words, applying your question:\n\n(x^3)/(x^7) = x^(3-7) = x^-4\n\nor 1/(x^4)
784622
16x^2 - 4x - 2\n= (4x)^2 -(4x)- 2\n= {(4x)^2 - 1} - (4x+1)\n={(4x+1)(4x-1)} - (4x+1)\n=(4x+1)(4x-2)\n= 2*(4x+1)(2x-1)
338855
You are looking for values of x such that\nx^3 > 4x^2  Solve for x by bringing everything to one side.\nSo subtract 4x^2 from both sides\nx^3 - 4x^2 > 0\nFactor x^2 out\nx^2(x-4) > 0\nx^2(x-4) = 0 when x = 0 or x = 4\nSince you're looking for positive numbers, you only have to test a number between 0 and 4, and then another number greater than 4\nAny number between 0 and 4 will result in a negative value.\nAny nummber greater than 4 will result in a positive value.  You are looking for the positive values because  the expression is greater than zero.\nThus, any number that is greater than 4 will give you the result that the cube of a number is greater than 4 times its square.
117698
so that x=4+ or - 3i\nthen transpose 4 to left\nx-4=+/- 3i square both sides\nthat's x^2-8x+16=-9\ntranspose -9 to left side\nx^2-8x+25=0
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
921629
You can easily solve this by realizing (4m - 3n)(4m + 3n) has a pattern (a + b)(a - b), which equals to a^2 - b^2\n\nSo,\n(4m - 3n)(4m + 3n) = (4m)^2 - (3n)^2\n= 16m^2 - 9n^2
933621
let's say u=4^x\nso the equation becomes:\nu^2-20u+64=0\n(u-16)(u-4)=0\nu=16 or u=4\n\nsubstitute 4^x in to get:\n4^x=16 or 4^x=4\n\nthus, x=2 or x=1
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
699632
Example:\n4^x = 8\n\nThe technique here is we change both sides into the same base.  Since we know that 4 = 2² and 8 = 2³, then we can change the equation into\n2^2x = 2^3\n\nThen, since the bases are equal, then the exponents must also be equal.  Thus,\n2x = 3\n\nTherefore,\nx = 3/2.\n\n----------------------------\nThe thing here is to change both sides into the same base.\nAnother example:\n81^[x(x + 2)] = 1/3^(9 - 20x)\n\nWe change both bases: 81 = 3^4, 1/3 = 3^-1.  Thus,\n3^[4x(x + 2)] = 3^[-(9 - 20x)]\n\nWe then equate the exponents:\n4x(x + 2) = (20x - 9)\n\nWe distribute:\n4x² + 8x = 20x - 9\n\nWe transpose everything to the left\n4x² - 12x + 9 = 0\n\nWe factor out\n(2x - 3)² = 0\n\nSquare root both sides\n2x - 3 = 0\n\nWe transpose -3\n2x = 3\n\nDivide 2.  Therefore,\nx = 3/2\n\n--------------------------\nHere's another example:\n\n2^x = 3\n\nI know that it is VERY DIFFICULT to change 3 into a power of 2.  Here, we use another tool: the logarithm.  If you are already ready, you can inform me so that I can teach you that.  Please feel free to contact me.\n\nkevin!\n^_^