7^(2x + 1) = 3^(x + 2)?
350639
7^(2x + 1) = 3^(x + 2)\nlog(7)(x + 2) = log(3)(2x + 1)\nIn the second step, you switched the x+2 and 2x+1\nLine 2 should be\nlog(7)(2x + 1) = log(3)(x + 2)\nContinue solving as you did before and you will get the correct answer.\n\nAlso, you wouldn't write ln9 just because it is not in simplest form\nln(9) = ln(3^2) = 2ln(3)\nSimilarly for ln(49) = 2ln(7)
1133153
7 - 2[4-6+2x] - 8+ x=6\n\n7 - 2[-2+2x] - 8 +x =6\n7 + 4 - 4x - 8 + x =6\n3 - 3x = 6\nx = -1\n\nDon't listen to the math goober - plug the value in and check it.
959661
if by this you mean\n\nx^2 - 2x - 2 = 0\n\nx^2 - 2x = 2\nx^2 - 2x + 1 = 3\n(x - 1)^2 = 3\nx - 1 = sqrt(3)\nx = 1 ± sqrt(3)\n\n-------------------------\n\nbut if you mean\n\n7x^2 - 2x - 2 = 0\n7x^2 - 2x = 2\nx^2 - (2/7)x = (2/7)\nx^2 - (2/7)x + (1/49) = (63/49)\n(x - (1/7))^2 = (9/7)\nx - (1/7) = (3/7)sqrt(7)\nx = (1/7) ± (3/7)sqrt(7)\nx = (1/7)(1 ± 3sqrt(7))
779926
Just expand the brackets\n\nf(x)=(3x^3+3x)(x+7)\n\nf(x)=3x^4+21x^3+3x^2+21x\n\nThen differentiate each single term by multiple the coefficient by the power and then decreasing the power by 1.\n\nf'(x)=12x^3+63x^2+6x+21
233929
it simplifies to 2x^n=x^n+1 \n\ntake the log of both sides (log x)\n\nlogx^2n=logx^n+1\n\nnlogx^2=(n+1)logx \n\nn&#92;(n+1) =logx/logx^2\n\nn/(n+1) = 1/2\n\n2n=n+1\nn=1\n\nPlug n back into equation\n\nx^1 +x^1 = X^2\n\nx+x=x^2\n2x=X^2\nx^2-2x=0\nx(x-2) = 0\nx=0 or x=2
267923
3^x=2^x/2\n\n(1.5)^x=0.5\n\nx=-1.71
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'
381253
3^(2x) + 3^(x) - 2 = 0\n(3^(x))^2 + 3^(x) - 2 = 0\ny = 3^x\ny^2 + y - 2 = 0\neither y = -2 or y = 1\n\ny = 3^x, then y > 0, then y = 1\n3^x = 1\n\nx = 0
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
919015
(8X^6/7) * (56/24)\n8X^6/3\n\n(Hint:  X^m/X^n = X^(m-n)
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
234025
(m^(2n))/(3*(a^x)) = 37       \n(3a^x)(37) = m^(2n)            *****cross-multiply\n111a^x = m^(2n)                *****3*37=111\na^x = (m^(2n))/111             *****divide by 111\nx= log (base a) (m^(2n))/111   *****take log (base a) of both sides\n\nOne note: If you meant the entire expression (3a) to be raised to the x power, you need to use parentheses around it. Hope that helps.
772527
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
293335
8+9+6+(1+2+3+4+5+7)^0 = 24\n\nX^(0) = 1
593331
x + 3 = x (x + 3)\n\nx + 3 = x^2 + 3x\n\nx^2 + 2x = 3\n\nx = 1 or -3
926657
(3 a^-7 b^7) / (9 a^7 b^-5) = ((3)(1/a^7)(b^7)) / ((9)(a^7)(1/b^5)) =\n\n((3 b^7) / a^7) / ((9 a^7) / b^5) = ((3 b^7) / a^7) times (b^5 / (9 a^7))\n\n= (b^12) / (3 a^14)    OR    3^-1 a^-14 b^12
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