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solve for x?

e^x - 3x^2 = 0

Solve for x.

(I know the answer, but am stuck on how to get there algebraically rather than graphically.)

Thanks!

6 Answers

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  • 1 decade ago
    Favourite answer

    What the hell? Are you a math undergraduate? If not, you shouldn't be seeing hairy equations like this. The solution to this involves a very nasty thing called the Lambert W function. There is no closed-form solution to your question.

    There exist three solutions to the problem exp(x) = 3x^2. Numerically, they are 0.910008, 0.458962 and 3.73308.

  • Anonymous
    4 years ago

    To be solved: 4x(x-7)-5x(x-6)=-3 STEP a million: improve 4x(x-7) - 5x(x-6) = -3 4x^2 - 28x - 5x^2+30x = -3 STEP 2: SIMPLIFY rather carry mutually and upload the like words (4x^2 - 5x^2) + (-28x + 30x) + 3 = 0 -x^2 + 2x + 3 = 0 Now this could properly be a quadratic in x STEP 3: resolve -x^2 + 2x + 3 = 0 you need to locate the aspects that would provide you a fabricated from -3 and addition of two i.e., x1*x2 = -3 (a million) x1+x2 = 2 (2) the place x1 and x2 are your strategies To on an identical time fulfill equations (a million) and (2) x1 = 3 and x2 = -a million now to envision x1*x2 = 3*-a million = -3 x1 + x2 = 3 + (-a million) = 2 This now solves your equation for x as x = 3 or -a million

  • 1 decade ago

    e^x - 3x^2 = 0

    e^x = 3x^2

    Taking log on both sides,

    log e^x = log (3x^2)

    x log e = 2 [ log 3x]

    x log e = 2[ log3 + logx]

    x= 2 log 3 + 2 logx

    x =2(0.4771) + 2 logx

    x= 0.9542 + 2 log x

    Differentiating wrt x,

    d/dx( x) = d/dx(0.9542) + 2 d/dx(logx)

    1= 2( 1/x)

    x= 2

  • 1 decade ago

    Aha!!!

    Go to http://www.hostsrv.com/webmab/app1/MSP/quickmath/0...

    And copy and paste e^x - 3x^2 = 0 into the solve box

    Click solve and gives the whole method

  • maussy
    Lv 7
    1 decade ago

    e^x=3x^2

    now use the ln (neperaian logs)

    x= ln3+2lnx

    sorry after you must use graphics

  • 1 decade ago

    Try using the solver on your graphing calculator if you can use one.

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