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2^x=x^2+7 Find x

  1. Oct 9, 2006 #1
    2^x=x^2+7
    Find x.

    The answer is 5. I have tried substituting 2^x=a, but that doesn't help. Is differentiation method required here? If not, how should I approach the question?
     
  2. jcsd
  3. Oct 9, 2006 #2

    dextercioby

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    The equation is awfully transcendetal. My advice is to find the solution by a graphical method.

    [tex] x=\frac{1}{\ln 2}\ln\left(x^{2}+7\right) [/tex]

    Daniel.
     
  4. Oct 9, 2006 #3
    2^x = x^2 + 7
    2^x - x^2 - 7 = 0

    Now use the Newton-Raphson method to find root(s) of f(x) = 0

    If the eqn, 2^x=x^2+7, looks simple enough you could try simple substitution.

    x=1: 2 = 1 + 7 -- nope, lhs too small
    x=2: 4 = 4 + 7 -- nope, lhs too small
    x=3: 8 = 9 + 7 -- nope, lhs too small
    x=4: 16 = 16 + 7 -- nope, lhs too small
    x=5: 32 = 25 + 7 -- yep, lhs = rhs

    If the above substitution method didn't work, and you went from lhs too small to rhs too large, with a value of x = xo, say, then you could try the Newton-Raphson method with xo as your first approximation.
     
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