are there any values of a that make a^m * a^n= a^mn. If so what are they.?
813081
a=1, a=0\n\nTO ANSWER YOUR EMAIL QUESTION:\n\na^m * a^n = a^(m+n)\n\nm+n ≠ mn for all m's and n's\n\nTherefore we need to look at the base, a.\n\nSince 1 to any power is always 1, we can state for certain that 1^m * 1^n = 1^mn for any m and n.\n\nSimilarly, since 0 to any power is always 0, we can state for certain that 0^m * 0^n = 0^mn for any m and n.\n\nHope that helped explain it a bit!
710821
I'm sure it isn't correct, though I'm not totally sure what you've done.  It looks to me as if you've raised both sides of \naⁿ + bⁿ= cⁿ\nto the power 1/n.  The trouble is, when you do this the left side is not equal to\n(aⁿ)^(1/n) + (bⁿ)^(1/n)\n\nYou can see this if you consider\n(5^3 + 2^3)^(1/3)= 133^(1/3), which is just slightly over 5, definitely not equal to\n(5^3)^(1/3) + (2^3)^(1/3)\n= 5 + 2\n= 7
921629
You can easily solve this by realizing (4m - 3n)(4m + 3n) has a pattern (a + b)(a - b), which equals to a^2 - b^2\n\nSo,\n(4m - 3n)(4m + 3n) = (4m)^2 - (3n)^2\n= 16m^2 - 9n^2
489740
let 0=0\n   then a^2-a^2=(a-a)(a+a)\n  then a(a-a)=(a-a)(a+a)\nthen a=2a\nthen 1=2\nthen 2=3
495003
a+a^20 DOES NOT =a^21. Don't confuse how to add and subtract exponents with real numbers.  a+a^20= a+(a*a*a*...*a)--where there are 20 a's inside the parentheses. -a+a^20 and a+a^20 are equal to a(-1+a^19) and a(1+a^19) respectively, using the distributive law.  Maybe this helps you see the difference, you can't turn that + sign into a * sign.  Lastly.  Working with exponents.  a^3*a^4= (a*a*a)*(a*a*a*a)=a^(3+4)=a^7.  That's the way it works, multiply "a's", and add the exponents of "a".  Now to multiply exponents, we take for example a^3 and raise it to the 4th power like this: (a^3)^4=(a^3)*(a^3)*(a^3)*(a^3)=a^(3*4)=a^12. I hope this helps.
697172
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
919306
3m^1/6 n^1/3 over 4n^-2/3 all to ^2\n= ( 3m^1/6 n^1/3 / 4n^-2/3 )^2\n= (3 m^1/6 n^{1/3+2/3} /4 )^2\n= (9/16) m^{2/6} n^2\n= (9/16) m^{1/3} n^2 '
1010848
First of all, u do not need Lagrange's identity to prove it. u just need to expand everything.\n\nSo, u want to use Lagrange's identity, fine. set a^m=a^2, a^n=x^2, substitute these into the Lagrange's identity and u get exactly [a^{(m+n)/2}+b*y]^(2) + [y*{a^(m/2)} - {a^(n/2)}*b]^(2) without even needing to do anything else.\n\nThis is not called proving!
4935
ur second equation is wrong. \n2a is not equal to a-squared (which u symbolized by ab, since a=b).
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
164104
maybe :\n\nKa = 1.3*10^-2
133829
8x^7\n\nThe first reference refers to this as the "Power Rule".\n\n  x^n'=nx^{n-1}
928756
m^6-n^6=(m^3+n^3)(m^3-n^3)\n(m^2-n^2)(m^4+m^2n^2+n^4)\nm^4-n^4=(m^2+n^2)(m^2-n^2)\nplugging in\nm^6-n^6/m^4-n^4\n=(m^4+m^2n^2+n^4)/(m^2+n^2)
8585
Hi,\n\nMagnetic Quantum Number "m" in relation to Azimuthal Quantum Number "l" can be depicted by- \n\nm = 2l + 1\n\nSo, for any subshell (defined by the value of L), there are (2L + 1) values of M possible given by-\n\nM = -L, -(L-1), ......, 0, ....., (L-1), L\n\nSo, from your given options, I think the second one- -L, 0, L  is the closest.
446810
Let r be an element if R, and hence in L.  Since L is an algebraic extension of K, all elements of L, there is an irreducible polynomial:\n\n    P(x) = a0 + a1x + a2x^2 + ... anx^n\n\nwith the ai's in K such P(r) = 0.\n\nSubtracting a0 from both sides, we see:\n\n    -a0 = r ( a1 + a2r + a3r^2 .. + anr^(n-1))\n\nDividing both sides by -a0 we get (a0 is not zero because P is irreducible):\n\n      1 = r (-a1/a0 - a2/a0 r - a3/a0 r^2 ...)\n\nThe right side is in R since the the coefficients -ai/a0 are in K and r is in r.  So we have an inverse of r in R.
810638
b=a+1\nc=a+2\nd=a+3\ne=a+4\nf=a+5\ng=a+6\nh=a+7\na^2+d^2+f^2+g^2\n=a^2+(a+3)^2+(a+5)^2+(a+6)^2\n=4a^2+28a+70\nb^2+c^2+e^2+h^2\n=(a+1)^2+(a+2)^2+(a+4)^2+(a+7)^2\n=4a^2+28a+70\nhence proved
150867
Your question makes no sense. There is no equal sign. What do you want to solve for? Why so many powers (a^b means a to the power of b)?