3^x * 9^-x = 27^x \n\n1/e^-x^2 = (e^3x) 1/e^2 \n\n2^2x + 2^x+2 -12=0\n\n2^x+1 = 5^1-2x \n\nln(x-3)-lnx=3 \n\nthanks for the people who helped dont have to answer all of em
138830
When you go through this, make sure I have put the parentheses in the correct place in the question.\n\n#1.\n\n3^x * 9^-x = 27^x \n3^x * 3^-2x = 3^3x\n3^-x = 3^3x\n-x = 3x (by taking log base 3)\nx = 0\n\n#2.\n\n1/(e^-x^2) = (e^3x)(1/e^2) \ne^2 = (e^-x^2)(e^3x)\ne^2 = e^(-x^2+3x)\n-x^2+3x=2 (by taking ln)\nx^2-3x+2=0\n(x-1)(x-2)=0\nx = 1 or x = 2\n\n#3.  (Notice that atheerai go this one wrong.  Make sure you check your answers by plugging them back into the original equation.)\n\n2^2x + 2^(x+2) -12=0\n(2^x)^2 + (2^2)(2^x) - 12 = 0\nor, alternatively\ny^2 + 4y - 12 = 0 where y = 2^x\n(y + 6)(y - 2) = 0\ny = 2^x = -6 is not possible\nbut y = 2^x = 2 gives x = 1\n\n#4.\n\n2^(x+1) = 5^(1-2x) \n(x+1)ln2 = (1-2x)ln5\nxln2 + ln2 = ln5 - 2xln5\nx(ln2 + 2ln5) = ln5 - ln2\nx = (ln5 - ln2)/(ln2 + 2 ln5)\n\n#5.\n\nln(x-3) - lnx = 3\nln((x-3)/x) = 3\n(x-3)/x = exp(3)  (taking e to the power of both sides)\n x - 3 = x*exp(3)\nx(1-exp(3)) = 3\nx = 3/(1-exp(3))
370366
3√x times 3√x, this means you multiply 3√x to 3√x\n3√x * 3√x = (3*3)*(√x * √x)\n                = (9)*((x^(1/2)) * (x^(1/2)))\n                = (9)*((x^[(1/2)+(1/2)])\n                = (9)*((x^[1])\n                = (9)*x\n                = 9x\nWhich is not 9√x.\n\nI hope this makes clear.
402560
1/2=x\n1/3x=9\n9/(1/3)=27\nx=27\n27-9=18\n18 is your answer
363106
1.\nln(5x + 2) + ln x = ln e\nln(5x + 2) + ln x = 1\ne ^ (ln(5x + 2) + ln x) = e ^ (1)\n(e ^ ln(5x + 2))(e ^ ln x) = e\n(5x + 2)(x) = e\n5x^2 + 2x - e = 0\nx = (-2 +/- sqrt(4 - (4)(5)(-e))) / 2(5)\nx = (-2 +/- sqrt(4 + 20e)) / 10\nx = (-2 +/- (2)sqrt(1 + 5e)) / 10\nx = (-1 +/- sqrt(1 + 5e)) / 5\ntest answers to ensure ln of positive number\nx = (-1 + sqrt(1 + 5e)) / 5\n\n2. \nln(x + 1) / ln(x - 1) = 2\nln(x + 1) = (2)ln(x - 1)\nln(x + 1) = ln((x - 1)^2)\nx + 1 = (x - 1) ^ 2\nx + 1 = x^2 - 2x + 1\n0 = x^2 - 3x\nx(x - 3) = 0\nx = 0 or x = 3\ntest answers to ensure ln of positive number\nx = 3
623121
Check the problem again.  It doesn't simplify much ... I would have expected the terms with e^-x to either both be subtracted or added to one, not having the diferent signs you show.\n\nIt's not zero even how you have it written.  If x=0, the result is (1+1)(1+1)- (1+1)(1-1)/(1+1)^2.  Or 4 - 0 = 4.  The answer isn't 4 either (for all x).
95569
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
368889
If u = e^x, then 1/u = 1/e^x = e^(-x)\n\n(u - 1/u) / (u + 1/u) = -t\n\nMultiply both sides by (u + 1/u)\n\nu - 1/u = -t (u + 1/u) = -ut - t/u\n\nMultiply both sides by u\n\nu^2 - 1 = -u^2t - t\nu^2 + u^2 t = u^2(1 + t) = 1- t\nu^2 = (1 - t)/(1 + t)\nu = +/-sqrt[(1-t)/(1+t)]\n\nRe-substitute u=e^x:\n\ne^x = +/-sqrt[(1-t)/(1+t)]\n\nTake ln of both sides:\n\nx = ln[sqrt[(1-t)/(1+t)]\n\nSorry I screwed that up the first time.  I need sleep.  Thanks for getting it right, Sidarth.
342825
this is a mixture of product power and chain ..simple but takes time\n((x^2+1)^9*(36x^4+77x^2+1).....\ne^(9x^2))/(2sqrtx)\nsry it cut my answer in half\ni didnt have to use ln\nmine is right i checked it on my calculator
449545
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'
549592
4^x-2^x-12=0\n(2^x)^2-2^x-12=0\nset 2^x=y\ny^2-y-12=0\n(y-4)(y+3)=0\ny=4 or -3\n2^x=2^2\nx=2\nignore the x=-3
512094
I assume the 9 is the base of the logarithm.  The equation becomes\n\nx=9^(5/2)\n\nsince 9=3^2\n\nx=(3^2)^(5/2)=3^5=243
328715
3^x = y\n9(3^x) = 9(y)\n(3^2)(3^x) = 9y\n3^(2+x) = 9y\nAnswer: c) 3^(2+x)
408104
4n^3/2 = 8^-1/3 Bring it to power 6\n4^6 x n^9 = 8^2\n(2^2)^6 x n^9 = (2^3)^2\n2^12 x n^9 = 2^6\nn^9 = 1 / 2^6 Take 9th root\nn = 1 / 2^(6/9) = 1 / 2^(2/3) =third root of (1/4)\n\nTh
529410
ln e^m = m\n\ne^(ln x^4) = x^4 as e^(in z) = z\n\nlog (base 2) 8 = 3 as 2^3 =8
407846
For the first one, 27 = 3^2*3, so sqrt(27) can be written as 3sqrt(3).  So sqrt(27) - sqrt(3) is 2sqrt(3).  Similarly, 12 = 2^2*3, so sqrt(12) is 2sqrt(3).  2sqrt(3) + 2sqrt(3) = 4sqrt(3).\n\nFor the second one, start by subtracting 2 from each side.  You get sqrt(3x-2) = 3.  Then square both sides.  That gives 3x-2 = 9.  Add 2 to each side:  3x = 11.  Divide by 3:  x = 11/3 = 3 2/3.
528099
Problem #1:\n2^(3x) = 4\n\nSince you have powers of 2, I would take the log base 2 of both sides:\nlog[2](2^(3x)) = log[2](2²)\n3x = log[2](2²)\n3x = 2\nx = 2/3\n\nProblem #2:\n5/6x = 10\n\nMultiply both sides by x\n5/6 = 10x\n\nDivide both sides by 10:\n5/60 = x\n\nReduce:\nx = 1/12\n\nYour answers are:\n2/3 and 1/12
342677
(2^x-2^-x) / 3 (the whole thing is divided by 3)\n\nThat is not an equation to be solved.   So I don't know where you are getting an equals sign (=).??\n\nYou may be able to simplify it, but you cannot say x= something unless you were provided with an EQUATION.\n\nI will do some work on it.\n\nAfter further review, try to input your value for x into the statement\n(2^x-2^-x) / 3 and what you get is another question.   Your value then EQUALS what?   Does this answer your question?    All you will get is a value.