why does a number raised to the power of zero equal one?
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
1032089
discovery of the number 0?
3940
1. It's one multiplied by ten to the twenty-third power.\n  -Otherwise ten to the twenty-third power.\n\n2. It is also an example of multiplicative identity.\n\nSorry if that is not what you are looking for.
999313
0^n = 0, where n is not equal to zero\nn^0 = 1, where n is not equal to zero\n0^0 = undefined. why?\n\nExpress the exponent 0 into n - n (n is not equal to zero)\n0^0\n= 0^(n - n)\n\nUse the property b^(x + y) = b^x · b^y\n= 0^n · 0^-n\n\nUse the properties 0^n = 1 and b^-x = 1/b^x\n= 1 · 1/0^n\n= 1/0\n= undefined\n\n^_^
357910
hello!!!!!!!!!! \n1+1 in binary is 10! it's not ten but one zero!!!
745485
a raised number is an exponent. for example 2^2 is 4. the ^ symbol is used for "raised".  another example is what is 6 raised to the 2nd power? well 6x6 is 36. its the same thing
822813
Well the log is the inverse of an exponent right?\n\nLog base B (X) = N is equivilant to \nB^N = X\n\nYou want to find why Log 2 = .301 right?\nLet's turn it into an exponent.  I assume you're using log base 10 so we get:\n10^.301 = 2 (approximately) right?\nThis is because .301 is the exponent that you must bring 10 to the power of in order to get 2.  That's the best explanation I can offer.
1003407
Yes, it is equal to 10 to the 100th power, because of the fact that 10 to the power of anything is equal to 1 followed by an amount of 0's that's equal to whatever the exponent is. For example: 10 to the power of 10 is equal to 10,000,000,000 or a 1 followed by ten zeros. So in this case, one googol is : 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000. Bonus fact: a googol times itself equals a googolplex, or 10 to the 200th power, or 1 followed by 200 zeros.
90067
There's no answer actualy. O cannot be divided by 0. \nThe answer infinity is just an assumption when \nnumber really near to 0 divided by another number near to 0.\n\nSuch as 0.0000000000000000001 / 0.0000001\nAlthough the example doesn't really show number near to 0, but if you change to 0.00000000000000000....01, you will see the same result
497395
1.\nBecause e to the power minus infinity is 0.
885437
The last two digits are '25'. \n\n(This is the same for ANY integer power of 5 greater than 1).\n\nThe calculator won't give you the answer since it is not accurate enough.
709175
If you need a positive result, then y will only be a real number when x - 1 is >= 0.\n\nx - 1 >= 0\nx >= 1\n\nSo x must be greater or equal to 1.
370054
One reason to use exponents is that you can use rules for exponents for computation.  For example, a square root of x is x raised to the 1/2 power, the fourth root of x is x raised to the 1/4 power.  What is the square root of a number times its fourth root?  It is x raised to the (1/2 + 1/4) power or the fourth root of x cubed.  Radical signs don't allow this type of manipulation.
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
272752
One way to understand this is to think through the meaning of a divided by b. It is a number such that when you multiply by b, you get a. \n\nThus 1 divided by 0 is undefined, because there is no number such that when you multiply it by 0, you get 1.\n\nIn the same way 0/1 is equal to 0, as 1 times 0 is 0.
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...