how do i break powers...6^153 ....how to simplify?
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...
888747
I think you meant 3^0 + 3^1 + 3^2 + ... + 3^(n-1)\nNote the parenthese.\nIt reduces to (3^n - 1) / 2\n\na^0 + a^1 + a^2 + ... + a^(n-1) reduces to (a^n - 1) / (a - 1)
913289
You end up with a remainder, so it's not pretty, but you should be able to do long division and get:\n\n3x^2 - 4x + (16/3) - 23/(3*(3x+2))
725839
I think it is this:\n\nFirst multiply through the groups..\nso you get:\n6n^7-4n^6-16+4n^6+2n^7+15\n\nThen gather like terms\n\n=8n^7-1\n\n(The -4n^6 and the +4n^6 cancel each other out)\n\nI don't think you can go further than that.\n\nGood luck.
595479
a. 4.5 * 10^5\nb. 6.7 * 10^(-5)\n\nHope this helps for your homework now try to get some sleep if you don't have too much more!  Ha ... I should listen to my own advice ... I think I'm going to it is late ... kinda.  :)
819914
Remember - a negative exponent is an inverse.  If all you are doing is multiplying and dividing, life is easy.\n\n(3a^-7 b^7)/(9a^7 b^-5)\n\nFirst, move the negatives from where they are to the other side of the miniscis:\n\n= (3 b^7 b^5)/(9 a^7 a^7)\n\n(You can verify this by rewriting, for instance, a^-7 as 1/a^7 and simplifying, but I promise you this works.)\n\nNow:  3/9 = 1/3\n\nb^7 b^5 = b^12\n\na^7 a^7 = a^14\n\nSo, b^12/(3 a^14) and we're done.\n\nIf there were addition and subtraction in the problem, then it gets an order of magnitude harder.  Then, there is no substitute for rewriting the negative exponents as reciprocals and simplifying from there.\n\nHope this helps.
193438
the 6^2 will divide out, and then you have to multiply the 2 with all the exponents on the inside....oh and r^0 is just one....so i think the answer would be t^-10 \nPS i just changed this answer, realized i made some stupid mistkae :-)
640496
To solve this you raise everything inside the first parenthesis to the power of 2 (essentially multiply the exponent by the power it is to be raised to.  Thus (6x^2z^4)^2 becomes (36x^4z^8).  Now we can multiply like terms:  (4*36)(x^4*X^3)(Z^8*Z^2).  When multiplying exponents, simply add the 2 powers.  144x^7z^5\n\n (6x^2z^4)^2(x^3z^2)4=(36x^4z^8)(x^3yz^2)4\n                       =(4*36)(x^4*X^3)(Z^8*Z^2)\n                       =144x^7z^5
520679
When dividing rational expressions you flip the second one and multiply\n (2x^2 + 5x)/(9x^2 – 16) ÷ (2x^2-7x + 6)/(3x^2 -x -4)\n= (2x^2 + 5x)/(9x^2 – 16)*(3x^2 -x -4)/(2x^2-7x + 6)\nFactor each one, then you can cancel anything on the top with anything that is on the bottom.
246754
6y^2, this leaves 6y^2(1-2y+6y^2)
782532
there is no built-in at-the-nth-power operator in c++,\n\nsimply use \n#include <math.h>\ndouble z = pow(3,2);
765513
do u mean 6x^2(-9x^3) = -54x^5\nmultiply the constants and add the exponents
725157
assuming the numbers after the parenthesis and variables are exponents...\n(x-1)^(3/2) [6x(x-1)^2-x^2(x-1)^3]\njust add the exponents together from the outside to the inside- (3/2) + (2) = 7/2\n\n(x-1)^(3/2) [6x(x^2-2x+1)-x^2(x^3-3x^2+3x-1)]\n(x-1)^(3/2) [6x^3-12x^2+6x-x^5-3x^4+3x^3-x^2]\n(x-1)^(3/2) [-x^5-3x^4+9x^3-13x^2+6x]\nx(x-1)^(3/2)(-x^4-3x^3+9x^2-13x+6)\nokay, that took me for friggin ever to type on a regular keyboard. Please, seriously consider me for the best answer!
917052
16/24....it's really not that hard...you should be able to solve it on your own.\nps. you might want to simplify it down to 2/3.
720847
Use logarithms. If you haven't learned them yet, it just helps to solve equations with powers.\na = b^x\nlna = ln(b^x) = xlnb\nx = lna/lnb\nx = ln(10)/ln(32.6087)\nx = 0.6607928852\nYou can use your calculator to check. I guarentee that it works.
51664
1.35 x 10th\n\nI probably typed it wrong.  1.35 times 10 to the 10th power.
757415
1) yes\n2) get rid of x factor, so -(x^3-1)/(x^2+1)\n3) You lost a negative, (2x+3)/(3-2x)\n4) Remove y factor: 7x/2y\n\nSteve