solve the exponential equation e^5x=7?
829854
Use logarithms to bring exponents down where you can deal with them.  Since the natural log is base e, that's the way to go here.\n\ne^5x = 7\n\n5x = ln 7\n\nand solve from there.
921129
x = 7 or 3\n\nTake the square root of each side of your equation, to get the following:\n\n| x-5 | = 2\n\nNotice I used absolute value signs, because either 2 or -2 squared will give us 4.\n\nSo, x-5 = 2 or x-5 = -2\n\nx-5 = 2 → x = 2+5 = 7\n\nx-5 = -2 → x = -2+5 = 3\n\nSo x=7 and x=3 are both solutions to your equation.\n\nHope that helped!\n\n~ ♥ ~
599509
Rewrite the equation as follow:\n(2x+7){(x-7)+(2x-3)} -(2x-1)(3x+1)=-1\n(2x+7)(3x-10)-(2x-1)(3x+1)=-1\n(6x^2+21x-20x-70)-\n(6x^2-3x+2x-1)=-1 =>\n(6x^2+x-70)-(6x^2-x-1)=-1\n(note: x^2 means x to the power of 2)\nRe-group:\n6x^2-6x^2+x+x-70+1=-1\n              2x-69=-1 \n=> 2x=68 => x=34
821366
2.7*10^-5\n7.041*10^5
1174116
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
1152335
Solve,\n\n3x = 2y - 5 ... (1)\n\ny - 5x = -1 ... (2)\n\nRearrange (2), we get,\n\ny = 5x - 1 ... (3)\n\nSubstitute (3) into (1)\n\n3x = 2(5x - 1) - 5\n\n3x = 10x - 2 - 5\n\n3x = 10x - 7\n\n10x - 3x = 7\n\n7x = 7\n\nx = 1 ... (4)\n\nSubstitute (4) into (3)\n\ny = 5(1) - 1\n\ny = 4 \n\nTherefore x = 1 and y = 4.
527221
The answer is 2x+7 with a remainder of 32
972739
f(x)=-5(x)+7\nf0g=-5[-9x^2-4x+5]+7\n     =45x^2+20x-25+7\n     =45x^2+20x-18\nfog(-9)\n=45(-9)^2+20(-9)-18\n=3645-180-18\n=3447
527619
use ln\n\nln(5^t) = ln(7^(t-1))\n(t)*ln(5) = (t-1)*ln(7)\n(t)*ln(5) = (t)*ln(7)-ln(7)\n(t)*ln(5)-(t)*ln(7) = -ln(7)\nt*(ln(5)-ln(7)) = -ln(7)\nt*(ln (5/7)) = -ln(7)\nt = -ln(7) / ln(5/7)\nt = ln(7) / ln(7/5)\n\nt=~5.78
209966
Set up an equation:\n\n72 = .6 * X\n\n72/.6 = X\n\nX = 120
968624
1.  \n\n7x + 5i = 7 - 6yi\n\nIsolate x:  \n\n7x = 7 - 6yi - 5i\n\nSolve the right hand side in the proper form of complex numbers (a + bi)\n\n7x = 7 + i(-6y - 5)\nx = 1 + ( (-6y - 5)/7 ) i\n\nYour goal is to get some value for a (the number without the i) and some value for b (the number next to the i).  That's why I factored out the i so that we get the complex number proper form.\n\n2.\n\nThe quadratic formula for ax^2 + bx + c = 0 goes as follows:\n\nx = (-b +/- sqrt(b^2 - 4ac))/2a\n\nIn the above question, a = 2, b = 7, and c = 6.  All you need to do is substitute.  Note that "+/-" means "plus or minus", meaning you'll get two solutions for x.
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
932643
4x + 7 = -7\n4x = 14\nx = 14/4\nx = 7/2
738884
x^2-5x-14=0\n=>x^2-7x+2x-14=0\n=>x(x-7)+2(x-7)=0\n=>(x-7)(x+2)=0\nTherefore either x-7=0 or x+2=0\nIf x-7=0,then x=7,and if x+2=0,then x=_2\nThereforex=7 or -2 ans
330086
Completing the square:\n\nGet the x^2 term by itself\nx^2 + 7x/5 - 6/5 = 0\n\nMove the constant to the other side\nx^2 + 7x/5 = 6/5\n\nTake half of the constant for the x term and square it: half of 7/5 is 7/10, squared is 49/100.  Add this to both sides of the equation\n\nx^2 + 7x/5 + 49/100 = 6/5 + 49/100\nx^2 + 7x/5 + 49/100 = 169/100\n\nThe left side is a perfect square trinomial - factor it\n(x + 7/10)^2 = 169/100\n\nTake the square root of both sides and solve for x\nx + 7/10 = +/- 13/10\nx = -7/10 +/- 13/10\nx = 6/10 or -20/10\nx = 0.6 or -2
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
1187142
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.