Can anyone explain to me everything you know about exponents?
917190
When raising a power to a power, you multiply:\n\nx^2^3 = x^6\n\nWhen multiplying exponents with the same base, however, you add:\n\nx^2 * x^3 = x^5\n\nThe same rule applies to division, same base = subtract.\n\nx^2 / x^3 = x^(-1)\n\nI'm not quite sure what you mean by dividing exponents, but recall that exponential fractions are handled in a specific manner:\n\nx^(3/2) = SQRT(x)^3
702351
I think you are looking too much into it, but here we go. \n\nIf you have the equation x^2 and say we pick points x= -1, and 1. \n\nWhen you plug the x's into the equation you will get a positive answer. This is why whenever the exponent is even, it will always bounce off the x- axis, because there is no way for it to go negative.\n\nfor the second question, say we take x^2 and plug in the same points above.\n\n-1 ^ 3 gives you -1. This would mean that ever x = -# will have a negative y value which means it will cross the x- axis\n\nTry it with more equations and you will understand
102907
Let:\nx = any number not equal to 0.\nm = any number\nn = any number\n\nProperty of exponents:\n\nx^m/x^n = x^(m-n)\n\nTherefore:\n1/1 =2/2 =3/3 = x/x = x^n/x^n = x^(n-n) = x^0 = 1
857787
This cannot be simplified without guessing because the problem is ambiguos for the following reasons:\n\n1) You have an odd number of parentheses. You have one open parenthesis and two close parentheses.\n\n2) It is unclear what the exponents are. Following the u^  you should always enclose in parentheses what you intend the exponent to be. Exception is when it is totally clear as in u^5.\n\nOne cannot tell what part of SQRT 63ux^2 is the intended exponent of u. \n\nMy guess is that the expression should be (8u^(sqrt63))(ux^2) -\nx*sqrt(28u^5). If this guess is accurate, then:\n\n8u^(3*sqrt(7))(ux^2)-x*2u^2*sqrt(7) =\n2ux[4u^(3sqrt(7))x-2sqrt(7)]\n\nBut my guess is probably wrong. Hope I was not rude.
416859
There is a rule of exponents that says (x^a)(x^b) = x(a+b).  Use this in reverse:\n\n(3^2x) - 3^(2x - 1) = 18\n(3^2x) - [3^(2x)*3^(- 1)] = 18\n(3^2x) - (3^(2x))/3 = 18\n3(3^2x) - (3^2x) = 54\n2(3^2x) = 54\n(3^2x) = 27\nlog(3^2x) = log27\n2xlog3 = log27\nx = log27/2log3\n= 1.5
725770
If I understand you, the exponents are 3x+4, x, and 12, all to base 8. In multiplying the left hand side, add the exponents:\n(8 to the 3x+4) times 8 to the x equals 8 to the (3x plus 4 plus x).\nSo, 8 to the 4x plus 4 equals 8 to the 12.\n4x plus 4 equals 12\nx equals 2
103365
The term EXPONENT was introduced by Michael Stifel (1487-1567) in\n  1544 in Arithmetica integra.  He wrote, "Est autem 3 exponens ipsius octonarij, & 5 est exponens 32 & 8 est exponens numeri 256" (Smith vol. 2, page 521). \n\nThe Latin translates roughly as\n\n  But 3 is the exponent of that same eight, and 5 is the exponent of 32, and 8 is the exponent of the number 256.\n\nSee link below for more explanation
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
769357
It can be helpful here to express this in terms of complex exponentiation. Note that sin x = (e^(ix) - e^(-ix))/(2i) and cos x = (e^(ix) + e^(-ix))/2. So:\n\n(sin x)^3 * (cos x)^4\n(e^(ix) - e^(-ix))^3/(2i)^3 * (e^(ix) + e^(-ix))^4/2^4\n(e^(3ix) - 3e^(ix) + 3e^(-ix) - e^(-3ix))/(-8i) * (e^(4ix) + 4e^(2ix) + 6 + 4e^(-2ix) + e^(-4ix))/16\n-1/(128i) * (e^(7ix) - e^(-7ix) + e^(5ix) - e^(-5ix) - 3e^(3ix) + 3e^(-3ix) - 3e^(ix) + 3e^(-ix))\n-1/64 (sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x))\n\nThus, all that tedious mucking about in the complex numbers has allowed us to reduce a hard-to-integrate product of powers of trigonometric functions into a sum of very to simple integrate functions. Thus:\n\n∫sin^3 x cos^4x dx\n-1/64∫sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x) dx\n-1/64 (- cos (7x)/7 - cos (5x)/5 + cos (3x) + 3 cos x) + C\ncos (7x)/448 + cos (5x)/320 - cos (3x)/64 - 3 cos x/64 + C
643228
This is unclear.  Use the symbol ^ when using exponents.\nDo you mean?\n(6x10^12)^12 times (3.2x10^8) divided by [(1.6x10^7)^3 times 10^2]?
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
235721
(3x²)²(1/3y²)²=(9x^4)(1/9y^4)=(x^4)/(y^4)\n\nin many computer languages the "^" symbol is used for exponents.
555506
1.54 x 10^7\n\njust ignore the brackets since everything is multiplication.\nmultiply 77 x .02 = 1.54\nand add exponents becuase its multiplication 4+3 = 7\n\ntherefore 1.54x10^7
53602
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
698821
The symbol "^" means raised to the power of ___ , in other words, it shows what exponent is used.\n\nIt doesn't matter what the initial amount is, the increase of 1% will result in the same amount of increase after a certain number of repetitions... which can be multiplied by the initial amount to determine the net amount...and you must add 1 to the rate of increase because you are retaining the initial amount.\nso after 1 cycle you will have: \n1.01 X 33,000 = 33,330\n\n2 cycles will result in:\n1.01 X 1.01 = 1.01^2 = 1.0201 X 33,000 = 33,663.33\n\n3 cycles will result in:\n1.01 X 1.01 X 1.01 = 1.01^3 = 1.030301 = 33,999.933\n\nPV = Present Value\nFV = Future Value\nR = Rate expressed as a decimal, 1% = .01\nn = number of cycles\n\nFV = PV (1+R)^n\n\nFor values of n greater than four or five, a scientific calculator is a big advantage!
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.