I need to know the explanation of an advantage of rational exponents over the radical sign?
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.
298073
ok so when u divide  stuff like that, u just subtract the exponents.  \nr^-3 s^5 t^2\n---------------\nr^2 s t^-2\n\nso for the r, the new exponent would be r^(-3-2) = r^-5\nfor the s, s^(5-1) = s^4.  (s by itself is equal to s^1.)\nfor the t, t^(2-[-2]) = t^4\nso the new expression is  r^-5 s^4 t^4. \nand if you really wanted to, u could rewrite r^-5 as 1/r^5. cuz thatz how negative exponents work. (x^-2 = 1/x^2)\nso then the final answer would be s^4t^4/r^5.\nhope this helps. let me know if u need more help
607947
i^-1= 1/i^1  you can get rid of the negative exponent by changing it to the reciprocal.
292874
exponents. for example 2^2 would be 2 to the power of 2 which would be 4. 2^3 would be 8. I hope this helped.
959950
When multiplying terms with exponents you multiply the terms and add the exponents.\n\na)\n(-x^3)(-x^2) = x^5 because neg * neg = pos and 2 + 3 = 5\n\na term without an exponent is assumed to have an exponent of 1 therefore (x^5)(-x) = -x^6\n\nb)\n-2(-2) = 4, 4(-2) = -8, -8(-2) = 16
460547
This looks close enough to the techniques used to show that the roots are close to the critical line. The actual estimates are difficult, though, and nobody has been able to actually give a proof using these techniques (obviously). Go ahead and try it. Perhaps you can see through to a proof. If so, you will be famous. If not, you will have learned somthing.
1045033
Rational exponents are easier to combine when more than one appears in an expression.\n\nI would say that something like the following would be easier to solve with rational exponents:\nThe square root of x to the 5th power, all divided by the cube root of x to the fourth power.\nWhen I see this equation, the way to simplify is to change the radicals to rational exponents--x to the 5/2 power divided by x to the 4/3 power--which can be further reduced by subtracting exponents.\n\nSince my first step in that problem is to change the radicals to rational exponents, that seems to "prove" (in some sense) that rational exponents are easier to work with in this type of problem.\n\nSo my "short answer" is that rational exponents are much easier to work with when I am combining terms with multiplication and division.
383816
!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
996746
Using radicals, imagine the 4th root of x^5 multiplied times the 3rd root of x^4.  &#92;4/x^5 times &#92;3/x^4.\n\nYou cannot just combine them under the same radical.  However, when you rewrite them as exponents to fractional powers, you get:\n\nx^(4/5) * x^(3/4).  Now, you can use the addition of exponents to a common base.  How do you add 4/5 and 3/4?  you get the common denom of 20 and make it 16/20 + 15/20 and get:\nx^(31/20).  How could you have ever known by looking at the radicals that you would have ended up with the 20th root of x^31st power?
791694
The rational roots of any polynomial:\n\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) \n\nare in the form ±p/q, where p divides a(0) and q divides a(n).\nIn your case, a(0) = -1 and  a(n) = 1, so the only rational roots possible are ±1.\nSubstitute ±1 into the equation to see if either is a root. (neither is)\n\nExample:\nx^3 + 4x^2 + 2x - 4 = 0\nThe only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ).  Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.\n\n*****************************************\nProof by contradiction.\nLet the rational number p/q be a reduced root (i.e, q  and p are relatively prime integers) to the rational polynomial equation:\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0\nand such that either p does not divide a(0) or q does not divide a(n).  [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]\n\nThen we know that \na(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0\nmultiply both sides by q^n to get:\n\na(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p \n+ a(0)q^n = 0\n\nor a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.\nand a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.\n\nThis means \na(n)p^n = -q * f(a,p,q)\na(0)q^n = -p * g(a,p,q)\n\nBy the first equation, q must divide a(n) * p^n.\nBut q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)\nBut likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.\nTherefore no such p/q can exist, therefore the roots are irrational.\n\nQ:\nwhy shouldn't q divide a(n) or p divide a(0)?\n\nBecause otherwise if both did then the roots COULD be rational (p/q could be a root).  If thats the case you prove/disprove it by example.  In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1.  So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).\n\nThere is no general way to determine  by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.
303643
Radical numbers are ones you are taking a root of.  The square root of 2, written as sqrt(2) (really written with that thing that looks like a long division sign).\n\nYou can take higher roots of numbers - cube root, for example, would look like a square root of a number, but with a 3 in the upper left of the outside of the sign (the 2 for a square root is omitted, and it is understood that if there is a square root sign with no number in that top left corner, then it is a square root).
150881
There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
663667
It is true when both exponent & base are the same.\n\nOtherwise, they have to satisfy the formula:\n\na = r^(1/(r-1))\nb = r^(r/(r-1))\n\nwhere you can plug in any value for 'r' & get a, b - these are the two numbers we want.\n\n\nExamples:\n\n1. Suppose r=2,\n\na = 2^(1/(2-1)) = 2^1 = 2\nb = 2^(2/(2-1)) = 2^2 = 4\n\nAs you have shown yourself, 2^4 = 4^2.\n\n\n2. Suppose r=1.5,\n\na = 1.5^(1/(1.5-1)) = 1.5^2 = 2.25\nb = 1.5^(1.5/(1.5-1)) = 1.5^3 = 3.375\n\nYou can show using a calculator that 2.25^3.375 = 3.375^2.25\n\n\nP.S. Another very interesting property you will notice is that:\nb=(a*r) & b=(a^r)\n\nsimultaneously!
857476
When working with the simplification of radicals you must remember some basic information about perfect square numbers. \n\nYou need to remember:  \n\nPerfect Squares \n4 = 2 x 2 \n9 = 3 x 3 \n16 = 4 x 4 \n25 = 5 x 5 \n36 = 6 x 6 \n49 = 7 x 7 \n64 = 8 x 8  \n81 = 9 x 9  \n100 = 10 x 10 \n \n Radicals (square roots) \n = 2 \n = 3 \n = 4 \n = 5 \n = 6 \n = 7 \n = 8 \n = 9 \n = 10 \n \n \n\nWhile there are certainly many more perfect squares, the ones appearing in the charts above are the ones most commonly used. \n\n \n\nTo simplify means to find another expression with the same value.  It does not mean to find a decimal approximation.\n \n\n \n\nTo simplify (or reduce) a radical: \n\n1.  find the largest perfect square which will divide evenly into the number under your radical sign.  This means that when you divide, you get no remainders, no decimals, no fractions.\n\nReduce:  the largest perfect square that divides evenly into 48 is 16. \nIf the number under your radical cannot be divided evenly by any of the perfect squares, your radical is already in simplest form and cannot be reduced further.\n\nwrite the number appearing under your radical as the product (multiplication) of the perfect square and your answer from dividing.\n\n\n \n\n3.  give each number in the product its own radical sign.\n\n\n \n\n4.  reduce the "perfect" radical which you have now created.\n\n\n\n5.  you now have your answer.
100388
This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
474886
1)  All courses of action have some advantages and some disadvtantages, however slight.\n\n2) All compound entities are subject to decay and dissolution.\n\n3) e to the i*Pi power plus one equals zero.\n\nThe first two teach open mindedness and humility.\nhttp://mathforum.org/library/drmath/view/53865.html
454725
If you have a radical in the denominator of a fraction, you have to multiply that fraction by that radical (or some form of it) in order to rationalize it.  If you had 12/sqrt(2), you'd have to multiply that fraction by sqrt(2)/sqrt(2), which will give you 12sqrt(2)/2, or 6sqrt(2).  (You're multiplying by sqrt(2)/sqrt(2) because that reduces to 1 - you have to do this, or else you change the value of the fraction.)\n\nNow if you're fraction has a more complicated denominator, you still have to multiply by some form of the radical in order to rationalize.  Take 12/(1-sqrt(2)).  You can't just multiply by sqrt(2)/sqrt(2) like we did for the last one because your denominator will still have a radical: sqrt(2)[1-sqrt(2)] = sqrt(2)-2.  So instead of just using the sqrt(2), use the conjugate of the denominator.  The conjugate is basically the same denominator, but changing the sign of the second term.  Here, your denominator is 1-sqrt(2), so your conjugate will be 1+sqrt(2).  So multiplying your original expression by 1+sqrt(2)/1+sqrt(2) will give you:\n\n[12/(1-sqrt(2))] * [1+sqrt(2)/1+sqrt(2)] = 12(1+sqrt(2))/(1-sqrt(2))(1+sqrt(2))\n\nUsing FOIL on the denominator gives you\n=12(1+sqrt(2))/[1+sqrt(2)-sqrt(2)+(sqrt(2))^2]\n=12(1+sqrt(2))/(1+2)\n=12(1+sqrt(2))/3\n=4sqrt(2)