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Solve the equation for x

  1. Jun 26, 2006 #1
    solve the equation:
    2^x^2-3=4^x (2 to the power of x squared - 3)

    i cant get it to work
    ln2^x^2 - 3=ln4^x
    (x^2 -3) ln (2)=xln(4)
    i know it'd be -3(ln2)=xln4... but i don't know what to do with x^2 part.

    i did other questions like this but with 5^2x and 3^1+x, what do I do with the X^2 that gets me stuck. If someone knows what I do with taht could you please tell me, im sure i can figure it out from there.
  2. jcsd
  3. Jun 26, 2006 #2


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    Do you mean 2x2-3 = 4x or 2x2 - 3 = 4x?

    2x2-3 = 4x
    (x2 - 3)ln(2) = xln(4)
    ln(2)x2 - ln(4)x - 2ln(3) = 0

    It's a polynomial in x.

    2x2 - 3 = 4x...

    Well actually, do you mean 2(x2) - 3 ... or do you mean (2x)2 - 3 ...?
  4. Jun 26, 2006 #3


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    Probably a typo, the constant term in the polynomial is just -3.
  5. Jun 26, 2006 #4
    still a bit confused

    it was the first one, the one you were working out.
    since your multiplying ln2 by (x^2-3) wouldnt it be 3 ln2?
    So from ln(2)x2 - ln(4)x - 2ln(3) = 0
    you get (x-6)(x+2) so x=6, x=-2.
    I tried putting this numbers in the original but it doesnt work out, what am I doing wrong?
  6. Jun 26, 2006 #5


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    Oh yeah, 3ln(2). So you'd get:

    ln(2)x2 - ln(4)x - 3ln(2) = 0
    x2 - 2x - 3 = 0
    (x-3)(x+1) = 0
  7. Jun 26, 2006 #6
    just to expand on what's been said and offer another way of starting off...
    2x2-3 = 4x is a bit like saying
    2x2-3 = 22x
    2 raised to the power on the LHS is equal to 2 raised to the power on the RHS such that both exponents are the same, ie: x2-3 = 2x :smile:
    Last edited: Jun 26, 2006
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