solve for x:        5^x=.125?
550920
5^x=.125\nTaking log (base e) on both sides\n\n-> xlog5=log(0.125)\n-> x= log(0.125)/log5 = -1.291099302
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.
463098
Think as follows.\n\n1 is 5^1 = 5\n\n2 is 5^2 = 25\n\n3 is 5^3 = 125\n\n4 is 5^4 = 625\n\nSee the pattern?, For the nth row is 5^n.\n\nHence, for what you wanted,\n\n5 is 5^5 = 3125\n\nAnd buddy, I always think before answering... :)\n\nRegards
1145188
7.125\n\nA = .5(B)(H)
427245
You can only take the log of numbers that are positive, so\n\nx^2 + 6x + 5 > 0\n\nSo (x+5)(x+1) > 0\n\nSo x would have to be greater than -1 and less than -5.
954896
x^2 + -2x = 5   Isolate the x^2 and x terms.\n\nx^2 + -2x = 5   Divide both sides of the equation by 1, which is the a-coefficient.\n\nx^2 + -2x + (-2/2)^2 = 5 + (-2/2)^2   Add half of the b-coefficient squared. In this equation that value is (-2/2)^2, and can be simplified to 1.\n\n(x + -2/2)^2 = 5 + (-2/2)^2   Write the x^2 and x terms side as a perfect square binomial.\n\n(x + -2/2)^2 = 5 + 1   Simplify the other side (simplify fractions).\n\n(x + -2/2)^2 = 6   Simplify the other side (combine numbers/fractions).\n\nSQRT[(x + -2/2)^2] = {+,-}SQRT[6]   Find the square root of both sides.\n\nWhen you combine and solve you'll get\n\nx = -1 plus or minus SQRT[6]
780717
Divide the coefficient of the x term by 2:  -5/2\nThen square it:  (-5/2)^2 = 25/4\nAnswer: 25/4
987583
2^(x+1) = 5^(1-2x)\n\nlog 2^(x+1) = log 5^(1-2x)\n\n(x+1) log 2 = (1-2x) log 5\n\n(log 2 + 2 log 5)x = log 5 - log 2\n\nx = (log 5 - log 2)/(log 2 + 2 log 5)
780025
x ≤ 5\n\n-- - - - - - -s-
117989
do it like grade 5 - 1 see - \n[X-(1/5X)^2] sqrt\n    (1/5X)^2 \n     1/25 (2X) \n[X-1/25(2X)]sqrt\nX=4 , understand
825103
No, you add exponents when you multiply, so you would have\n5*(-4)r^(2-3) = \n-20r^-1 = -20/r
894640
.5K^(2/5) = .14K\nK^(2/5) = .28K\n(5thrt)(K^2) = .28K\nK^2 = (.28)^5K^5\nK^2 = 0.001721K^5\n1 = 0.001721K^3\nK^3 = 581.0451\nK = 8.345
720847
Use logarithms. If you haven't learned them yet, it just helps to solve equations with powers.\na = b^x\nlna = ln(b^x) = xlnb\nx = lna/lnb\nx = ln(10)/ln(32.6087)\nx = 0.6607928852\nYou can use your calculator to check. I guarentee that it works.
1068358
44\n\nthe cubed root of 125 is 5... this means the cube was 5x5x5. 8 of these are the coners (having 3 sides red) then that leaves 12 edges with 3 cubes no coners (2 sides red). \n\n8 + 12*3 = 44
1175698
cube root of (-125)=-5 \n\n3*sqrt(-125)=\n3*sqrt(25*5*-1)=\n3*5*i*sqrt(5)=\n15i*sqrt(5)\n\nnot sure which you were asking. For future reference, if you are talking cube or higher roots, use the n^(1/y) notation. A lot less confusing. sqrt is just for square roots.
640221
8 = 2^3\n.125 = 1/8 = 8^(-1) = 2^(-3)\n8^x = .125\n(2^3)^x = 2^(-3)\n2^(3x) = 2^(-3)\nWhen the bases are the same, then the exponents are the same.\n3x = -3\nx = -1
779760
if by this you mean\n\nf(x) = (x^(2/5)) * 5(x - 1), then\n\nf'(x) = (2/5)x^((2/5) - 1)*(5(x - 1)) + (x^(2/5))(5(x - 1)')\nf'(x) = (10(x - 1)/5)x^((2/5) - (5/5)) + 5x^(2/5)\nf'(x) = (2(x - 1))x^(-3/5) + 5x^(2/5)\nf'(x) = ((2x - 2)/(x^(3/5)) + 5x^(2/5)\nf'(x) = ((2x - 2) + 5(x^(2/5) * x^(3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^((2/5) + (3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^(5/5))/(x^(3/5))\nf'(x) = (2x - 2 + 5x)/(x^(3/5))\nf'(x) = (7x - 2)/(x^(3/5))\n\n------------------------------------\n\nBut if you mean\n\nf(x) = (x^2)/(5(x - 1))\n\nf'(x) = (x^2)(5(x - 1))^(-1)\nf'(x) = (2x^(2 - 1))((1/5)(x - 1)^(-1)) + (x^2)((-1/5)(x - 1)^(-1 - 1))\nf'(x) = ((2x)/(5(x - 1)) + ((x^2)/(5(x - 1)^2))\n\nANS : f'(x) = ((2x)/(5(x - 1)) - ((x^2)/(5(x - 1)^2)\n\nor you can do it like this\n\nf'(x) = (2x(x - 1) - x^2)/(5(x - 1)^2)\nf'(x) = (2x^2 - 2x - x^2)/(5(x - 1)^2)\nf'(x) = (x^2 - 2x)/(5(x - 1)^2)\n\nor\n\nf'(x) = (x^2 - 2x)/(5(x - 1)(x - 1))\nf'(x) = (x^2 - 2x)/(5(x^2 - 2x + 1))\nf'(x) = (x^2 - 2x)/(5x^2 - 10x + 5)\n\nANS : (x^2 - 2x)/(5x^2 - 10x + 5)