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Exponential simultaneous equation with powers

  1. Jun 1, 2010 #1
    Hi there,Struggling with this problem, anyone know how this might be solved?

    (e^(r-0.04)^(5/12))(0.9463) = (e^(r-0.04)^(11/12))(0.95152)

    and solve for r

    Ans: r~2.90%

    Would really appreciate it!
  2. jcsd
  3. Jun 1, 2010 #2


    Staff: Mentor

    It's hard to tell for certain what you're working with. Is this it?
    [tex]0.9463(e^{r - .04})^{5/12} = .95152(e^{r - .04})^{11/12}[/tex]

    If so, when you raise an exponent to a power, the exponents multiply, so the above can be rewritten as
    [tex]0.9463e^{(5/12)(r - .04)} = .95152e^{(11/12)(r - .04)}[/tex]

    Now take the natural log of both sides.
  4. Jun 2, 2010 #3
    Thank you Mark44.
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