^= to the power of\n-root\n\n\nsimplify 4^2x3-64\n         --------\n          16^7/2 x 2^0\n                        into the form 4^n where n E Z
129872
Here : \n\n4^2x3-64 reads 4 squared times the cube root of 64.\n\n= 4x4 x (64)^1/3 = 4 x 4 x 4 = 4^3 = 64\n\nAND \n\n16^7/2 x 2^0 = ((4)^2)^7/2 x 1 = (4)^ (2 x 7/2) = 4^7\n\nCOMPRENDE` ?
416398
If you are asking if the negative cubed root of 64 is 8, then no it is wrong.  The cubed root of 64 is 4, so your answer should be negative 4.   4*4*4 is equal 64, so taking the cubed root would be 4.     Taking the square root is like multiplying to the (1/2) power, so taking the cubed root is like multiplying to the (1/3) power.  So 64^(1/3) is equal to 4.  then the negative of that answer is -4.
429487
a^4 - b^4\n\n= (a^2 + b^2)(a^2 - b^2)\n\n= (a^2 + b^2)(a + b)(a - b)
503211
If your problem is:\n((((36x)^2y)^5z)^3), then you know that the whole thing is raised to a power of three, so the cube root is the expression without the three.\n\nIf the problem is like it is written in the person above's answer, then you indeed may do each part separately:\ncube root of 36 * cube root of x^2 * cube root of y^5 * cube root of z^3\n\nOf course, you'll end up with fractional powers. Just remember that when you take the nth root of something, it is equavalent to raising it to the 1/n power (the sqaure root of x = x^(1/2)). Hope this helps, and good luck!
335805
4^2=4^x+1\n2=x+1 (because the two powers must be equal\n2-1=x\nx=1
571066
I seem to be the first with the correct solution... the confusion is obviously the notation.  If you can't type superscripts easily, you can always use ^ to mean exponentiation.  (Or Alt-253 on a PC keypad will get you a superscript ²).\n\nThe solution is to simply substitute x in your equation with whatever is put inside the parentheses.  It can be confusing if that includes another equation, but it is still simple substitution.\n\nGiven:\nf(x) = 2(x^2) -1 \n\nf(2x) = 2((2x)^2) - 1\nf(2x) = 2(4(x^2)) - 1\nf(2x) = 8(x^2) -1\nf(-x) = 2(-x^2) - 1\nf(-x) = 2(x^2) - 1\n\nf(2x) - f(-x) = 6(x^2)\n\nf(x+1) = 2((x+1)^2) - 1\nf(x+1) = 2(x^2 + 2x + 1) - 1\nf(x+1) = 2x^2 + 4x + 2 - 1\nf(x+1) = 2x^2 + 4x + 1\n\nSo your answers are:\n1) 6x^2\n2) 2x^2 + 4x + 1\n\nWritten using Alt-253 for the exponent, the answers are:\n1) 6x²\n2) 2x² + 4x + 1
118004
work each one out...\n\na.) 2^6 + 1/2^4 = 64 1/16\nb.) (2^2+1/2^5)^-2 = when you take the reciprecal is going to end up being a small value so that's not it...\nc.)1/2^15.... the greater the value in the denominator the less the value is.\nd.) -(1/2^2 2^4)^-3 = yet again a reciprical makeing the obvious answer A since it's the only one that yields a number that isn't a fraction...
206019
x=-(-1)+/_ {(-1)^2 -4*5*12} all over 2*5\n= 1+/- {1-240} over 10\n= 1+/- i{249} over 10\n\n{}= square root, ^= to the power of,\n\nif you need more help let me know via email.
228277
2-root-3 is the same as 2 ^ 1/3 (^ is exponent sign)\n\n2 ^ 1/3 squared is (2 ^ 1/3) ^ 2 ... you multiply the exponents, so it equals\n\n2 ^ 2/3
340961
Remember how powers are defined. 2^3 = 2*2*2. So, a nice feature of exponents is that multiplying numbers adds the exponents = for instance, 2^3*2^4 = (2*2*2)*(2*2*2*2) = 2^(3 + 4) = 2^7\n\nNow, lets say you have the problem 2^7*2^x = 2^4. How would you solve this? By the nice exponent property above, this is the same as 7 + x = 4, so x = -3. But expanding these out, 2^7 = 128, 2^4 = 16, so 2^-3 should be 1/8.\n\nThis is also why anything (except 0) to the 0th power is 1:\n\na^x*a^0 = a^(x + 0) = a^x, so a^0 = 1
355719
2/√6 = √(2/3) sorry can't get rid of the fraction.\n\nBy the way, ^ means "raise to the power"
419237
1)  x^ 1/2 + 2/(x^1/2) - 1\n\nFirst, x^ 1/2 power = square root of x\n\nSq rt of x + 2/(Sq rt of x ) - 1\n\nTo get rid of the Sq rt of x in the denominator, multiply the fraction by Sq rt of x / Sq rt of x)\n\nSo:\nSq rt of x + 2 Sq rt of x - 1\n\nAdd the first two\n\n3 sq rt of x - 1\n\n2)  cube root 5(sq rt 5^3)\n\nI would put this into powers if possible:\ncube root of 5 = 5^1/3 x 5^1/2^3\n\nNow, the power rule:\n\na^m^n = a^mxn\nSo:  5^1/3 x 5 ^3/2\n\na^m x a^n = a^m + a^n  This means we will need like denominators:\n\n5^2/6 x 5^9/6 =25^(2/6+9/6) = 25^11/6\n\n3)  x^5/3 - x ^ 1/3 * z ^ 4/3\n     -----------------------------\n      x ^ 2/3 + z ^ 2/3\n\nAs before, attempt to combine as much as possible\n\nx ^ 4/3 x z ^ 4/3\n------------------\nx^2/3 + z ^ 2/3\n\nSimplify:\n[(x^4z^4)]^1/3\n----------------\n[x^2 + z^2]^1/3\n\nThe cube roots fall out\n\n(x^4z^4)\n------------\n(x^2+z^2)
80588
In the future, us ^ for exponents. It's pretty much standard for typed equations. But I digress....\n\n6. 10[(4 + 5).2. - 12)]\n10[9^2 - 12)]\n10[81 - 12]\n10 * 69\n690\n\n-------------\n\n9. 7.0.\n\nAny number raised to the zero will equal 1.\n\n-------------\n\n27. 3(2a + 4) + 2(4a + 2)\n6a + 12 + 8a + 4\n14a + 16\n\n-------------\n\n43. 32 - (-15) \n\nSubtracting a negative is the same as adding so...\n32 - (-15) = 32 + 15 = 47\n\n-------------\n\nX. Evaluate the expessions. b = 6, c = -3.\n2b - c\n\n2(6) - (-3) = 12 + 3 = 15\n\n-------------\n\nXIV. Sequences and tables. look at the table below and awser the questions.\n_____\nn ! t!\n1 ! 4!\n2 ! 8!\n3 !12!\n4 !16!\n5 !20!\n------\n\n67. Write and equation in the for of t = ____ that describes the table above.\n\nt is always 4 times n, therefore t = 4n\n\n-------------\n\nXIV. Sequences and tables continued. Look at the table below.\n_____\nn ! t!\n1 ! 4!\n2 ! 7!\n3 !10!\n4 !13!\n5 !16!\n-----\n69. Write and equation in the form of t = ____\nthat describes the table above.\n\nThis on is trickier but notice that the value of t increases at 3 times the rate of n. At that point, we know the 'slope' is 3. Then, the 'intersection' would have to be 1 to make the equations true. So....\n\nt = 3n+1
443269
You can start by factorizing 3x^4 from everything\nm(x) = 3x^4 (2x^3 + 5x^2 - 3x - 3)\n\nI think there's something wrong with your problem, none of the possible roots given by the Ruffini's rule are working. Perhaps, if instead of 6x^7 + 15x^6 - 9x^5 - 9x^4 it was 15x^7 + 15x^6 - 9x^5 - 9x^4, then it could be factorized very nicely in\nm(x) = 3x^4 (x+1)(5x^2-3) \nor if instead you wrote 6x^7 + 6x^6 - 9x^5 - 9x^4 then it'd factorized to\nm(x) = 3x^4 (x+1)(2x^2-3)
469334
you can try breaking the problem up into pieces:\n32/4 = 8 (these are just the coefficients)\ny^16/y^4 = y^12 (remember you subtract exponents if the bases are the same and being divided)\nso your final answer is:\n8y^12\n\n\nthere really are no steps! \nhope this helps!
127654
4x^3y^2-16y^3-9y=\ny(4x^3y-16y^2-9)
513482
2^(1-x)=2/2^x\nrewriting the equation,\n2^x+2/2^x-3=0\n2^2x+2-3*2^x=0\nsubstitute 2^x=t\nt^2-3t+2=0\n(t-2)(t-1)=0\nt=2 or 1\n2^x=2,x=1\n2^x=1, x=0\nso x=0 or 1