how do you find the apporximated value of a fractional exponent on paper?
152856
There are lots of methods for approximating roots by hand. \n\nFor example, here is a more general technique, called the Newton-Raphson method. \n\nLet F(X)=(X^2)-2\n\nYou want to find the place R where F(R)=0.\n\nSo to find the roots of F(X), iterate. \n\n\nXn+1=Xn-F(Xn)/F'(Xn)\n\nwhere F'(X)=2X\n\nPick a starting value for X\n\nX0=1, say, and then compute X1, then X2, and so on. \nIt should converge pretty rapidly. \n\nAnd it works for other fractional powers too !
539053
4.43x10(at the power of -4)
535461
To break it up into something easy for a computer to read, break the power down to something divisible by 2 and then split that. Here's an example:\nx^153 = x*x^152\nx*x^152 = x(x^76)^2\nx(x^76)^2 = x((x^38)^2)^2\nx((x^38)^2)^2 = x(((x^19)^2)^2)^2\netc.\nI'm not sure that's what you wanted to know, though...
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.
580659
Ok, first, its not 1/y^2.  It's 1/y^5.  \n\ny*y*y*y*y\n-------------\n(y*y*y*y*y*y*y*y*y*y)\n\n\n= 1\n-----------------\ny*y*y*y*y\n\n\n= 1\n----------\ny^5\n\nAnd you write it just like that so as to keep the exponents positive.
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).
812981
1/2^x < 0.01 \nthen multiply both sides by 2^x\n1 < 0.01 2^x\nnow divide both sides by 0,01:\n1/ 0.01 < 2^x\n100 < 2^x\n~
984623
It will be 1/(x^2)\n\nFor example\n\n(1/2)^2 = (1^2)/(2^2) = 1/ 4\n(1/3)^2 = (1^2)/(3^2) = 1/ 9\n(1/4)^2 = (1^2)/(4^2) = 1/16\n\nSo the result actually gets smaller even though you are squaring it
739921
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.
770136
P = Ae^rt\nP = 100e^(.2*52)\nP = 100 * 32824.21\nP = 3282421
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
985294
you just keep on deriving.\n\nI find it easier to split things into exponents.\n\nI am assuming your question goes\n\n(1/(2x))  - 4 = g(x)\nor\n\n(1/2)*x^-1     - 4 = g(x)\n\ndifferentiate once\n\ng'(x) = -(1/2) * x^-2 \ng''(x) = -2 * (-1/2) * x^-3  = x^-3\ng'''(x) = -3 * x^-4  =  -3/(x^-4)  **ans**\n\nPlug 3 into this... bah\n\nFor reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).\n\n\n===!using the correct g(x) which should be g(x) =1 / (2x -4)===\n\ng(x) = 1 / (2x -4) = (2x - 4)^-1  don't forget the chain rule\n\ng'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2    chain rule again\n\ng''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3    chain rule again\n\ng'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4\n\ng'''(x) = -48/[(2x -4)^4]\ng'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16   = ***real ans***\n\ng'''(3) = -3
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
728458
Here is how you do it (I'll use ^ for the power sign).\n- Divide by P:    (1+r/n)^n = F/P\n- Take the 1/n the power:     (1+r/n) = (F/P) ^ (1/n)\n- Solve for r:   r= n* ((F/P) ^ (1/n) -1 )\n\nIn this case F=20000, P = 12600, n=12 therefore r = 0.47 or 47%\n\nSeeing this answer I wonder if I didn't misinterpret your notation, but with the method given, you can solve the question for yourself.
258044
divide the exponent by two
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
738843
I'd recommend graphing the function on a calculator (I use a TI-83) and use the 'trace' to find zeros.