How to calculate x^x = 100 without using trial & error or brute forcing ?
398683
Use log you will find\n\nor put\n\nx=3.597285024\n\nIts done
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
42255
The web page [1] below gives the solution to\n\n    x^x = 1333\n\nas\n\n    x = e^(W(ln(1333)) = 4.669202878...\n\nwhere W is Lambert's W function (see web page [2] or more generally web page [3]).\n\nSo for x^x = y if y is in the appropriate range then x would be e^(W(ln(y))).
954164
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
136719
How about this way: x = 2.31717171717\n\n100x = 231.71717171\n\n100x -x = 99x = 231.71717171- 2.31717171 = 229.4\n\ntimes by 10 to clear decimal\n\n990x = 2294\n\ndivide by 990\n\nx = 2294/990\n\ndivide by 2 to reduce\n\n1147/495
731776
simple!!!\nthe answer is : 12x^2\nthe rule of addition intergers with exponents is simple.\nyou just add the whole number (that is 4+8) and then follow the integers (that is x^2)... get my point?\ni hope i have helped you with your assignment....
56508
first, do 2 to the 12th power (exponents take precedent over multiplication) which is 4096. Then multiply that by 86 and get 352256.  It's quite simple, really.
997010
x^0 is equal to  x^(1-1) = x^1/x^1 = x/x = 1/1 = 1\n0 is actually undefined. (0^0) because you're dividing by 0.
951594
Let x be one number\nother number=100+x\nproduct\n=x[100+x]\n=100x+x^2\ndifferentiating\nf'x=100+2x\nthis is a minimum \nwe put this=0\n100+2x=0\nx=-50\nthe numbers are\n50,-50
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.
174070
You need to isolate x by using your rules of logarithms.\n\nI'll get you started:\n\ntake the log of every term in the equation\n\nlog2^x+log3^(x+1)=log5\n\nnow all of your exponents can drop to be multiplication on the logs.\n\nxlog2+(x+1)log3=log5\nnow distribute and refactor putting the x's together\n\nxlog2+xlog3+log3=log5\nx(log2+log3)+log3=log5\n\nnow you have one x to work with and a simple algebraic problem
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...
193745
1.25 * 10^-3 * 8 * 10^-5\n\n(1.25 * 8) * (10^-3 * 10^-5)\n\n(10^1) * (10^-8)\n\nANSWER : 1.0 * 10^-7
283872
x^x^x^x..... like that upto infinity right\nok then we will take one x out from the  equation and the rest of them will be assigned the value 9\n\ni.e x^9=9\n\nx=9^1/9\n\nor x=1.27\n\nhope u understood
457868
It's possible to get 100 using even THREE eights!\nAnd this is in DECIMAL, not binary base!\n\n100 = -log_8 (log_8 (sqrt(sqrt(...sqrt(8))))),\n\nwhere log_8 is logarithm in base 8\nand the number of square roots is 300.\n\nLet's prove it.\n\nA = sqrt(sqrt(...sqrt(8))) {300 roots} =\n= ((8^(1/2))^(1/2))...^(1/2) {300 powers} =\n= 8^(1/2 * 1/2 * ... * 1/2) {300 multiples} = \n= 8^(1/8 * 1/8 * ... * 1/8) {100 multiples} = \n= 8^(1/(8^100)) =\n= 8^(8^(-100)).\n\nlog_8 (A) =\n= log_8 8^(8^(-100)) =\n= 8^(-100).\n\nlog_8 (log_8 (A)) =\n= log_8 8^(-100) =\n= -100.\n\nHence, -log_8 (log_8 (sqrt(sqrt(...sqrt(8))))) = 100.\n\nQED\n\nYou can use forth eigth, for example, to change every one 8 in that formula to sqrt(8*8).
211814
2x^2 +3x-6 well that is as simple as it can get unless you have a number for the variable x
875144
you have to take the square root of each side, then just simplyfy\n\nx = 1.2472191289247