What are the similarities between radicals and rational exponent notation.?
535859
all radicals can be expressed as rational exponents. the roots of all radicals be expressed as denominators of exponents. a regular exponent will be the numerator of the exponent.\n\na regular x has an exponent of 1,  x = x^1\n\nso, for example, the square-root of x (√x) can be rewritten as:\nx^(1/2)\n\nthe cube-root of x (3√x) can be rewritten as:\nx^(1/3)\n\nthe cube-root of x^2 (3√(x^2)) can be rewritten as:\nx^(2/3)\n\nand so on\n\nthis is why the square-root of x^2 = x, becuase\n= x^(2/2) = x^1 = x
1012665
if the multiplication involves the same radical, (squarel root, cube root, etc.) then that is equal to the product of the radicals\nie sqrt(5) * sqrt (2) = sqrt(10).\n\nif they are different radicals, then it is more difficult. the only way is to change the radical to the same radical through manipulation of the number/radical.\n\nfor example 4 * sqrt (2) = sqrt (4^2) * sqrt (2) = sqrt (16) * sqrt (2) = sqrt(32).
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!
1150969
a = 3^(1/3)\nb = 4^(1/3)\n c = a+b\n If you use the primitive element theorem  you will see that c is a primitive element for Q(a)(b)\nso Q(a)(b)=Q(c), which implies that the degree of the minimal polynomial is 3*3 = 9\n By primitive element theorem a is written as a rational fraction in c, a = f/g(c),\nso a^3 = f/g^3(c), or g^3(c)*a^3 = f^3(c), g and  f polynomials in Q.\n so the minimal polynomial will be 3g^3-f^3\n( hoping that g and f have at most degree 3 in c)\ng and f can be computed as in the proof of primitive element theorem\ni.e compute gcd (x^3-3, (c-x)^3-4))\n Maybe there is a better way...
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.
865479
First of all is the radical in the numerator over just the 10 or over the entire 10y^3?\n\nThe only this you can factor out and cancel is a radical 2 on the top and bottom.  Other than that, you can't really simplify this problem.  Are you sure you wrote it down correctly?
640665
radical is the same as root...\n\n"3 radical 6" = 3*sqrt(6)
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
574952
Use your logorithm rules to combine the terms on the lefthand side into one term\n\nlog (x^2/(8x)) = log (x-10)\n\nNow, if the logorithms are equal, the arguments must be equal\n\nx^2/(8x) = x-10\n\nCross-multiply\n\nx^2 = 8x^2-80x\n\n7x^2-80x=0\n\nx(7x-80)=0\n\nSolving this, you get x=0 or x=80/7. but you have to reject x=0 because it's not in the domain of the original expression (you can't have log(0)).\n\nso x=80/7
844451
to the right, i.e. x-> infinity,  3x^2 - 5x -7^-x is more like 3x^2,\nnow , to the left, ie x-> -infinity, 3x^2-5x-7^{-x} is more like the exponential -7^{-x} \nis this what you mean, may be if you provided an example, i could help you better.\n'
749268
1.45*780*132*10^9/825*555*155*10^-11\n4578*132*10^21/825*555*155\n\n\n2.115*10^9*688*10^5*9*10^3/\n256*10^5*75*10^3*245*10^2*92*10*-5\n=115*688*9*10^17/256*75*245*92*10^5\n=115*688*9*10^12/256*75*245*92
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.
326552
Sometimes they're just easier to write out.  "The fourth root of x raised to the 9th" in radical form is a 4 on the outside and an x^9 inside.  But in decimal form, it's x^9/4.  For some, drawing that radical sign is a pain and looks sloppy after a while, whereas fractions are easier to write and understand (no confusing that with a division sign!).\n\nOther times it's easier to work with mathematically.  Consider:\n\ncuberoot(x)/x^4\n\nThat doesn't look like it's simplified as is, but in exponent form, you can do more to it.  cuberoot(x) = x^(1/3), so\n\nx^(1/3)/x^4 = x^(-11/3)\n\nExponents vs. radicals is great when multiplying or dividing numbers or variables with the same base.  If you're multiplying, you simply add the bases:\n\nsqrt(x) * cuberoot(x) = x^1/2 * x^1/3 = x^5/6\n\nAnd if you're dividing, as in my first example, you subtract the bases.
391062
It's an easy equation . The solution is: \nX=(ln (sqrt5-1)/2)/ln(2/3)....sqrt means the radical sign.
233017
the number of roots for any equation is the degree of the equation. here it is 3 (coz the max power or exponent is 3)
141990
(assuming the second 2 is an exponent) 2*2*2*7*5*g*h*h \nyou really should learn to do this if you are ever going to get through school though
963683
there are 3 cases for a number A raised to the power of n:\n\n1) n greater than or equals to 1.\nWe simply say that "A is raised to the power of n"\n\n2) 0 < n < 1 and n is a rational fraction expressed as 1/m, m is an integer\nWe say that "A is raised to the mth root"\n\n3) n is any rational fraction not included in case 2, expressed as p/q, p and q are integers\nWe say that "A is raised to the power of p and qth root" which therefore involves both power and root\n\nAnd you're right about radicals being the sign used. By right, for square roots (which is actually 2nd root), we should add a "2" on the top left hand corner of the radical sign, but by default a radical without any number is taken as square root. But for all other nth roots, the radical sign must be accompanied by the number n.