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24 = 10(e^x/20 + e^-x/20)

  1. May 24, 2012 #1
    I'm reviewing for calculus and I've been stuck on this problem for a couple days. Can't seem to figure it out. I keep coming up with ln2.4 = x - x which is just silly, or x = √(400ln2.4) ≈ 18.7. I know the answer is x ≈ 12.5. But I can't figure out the correct process.

    1. The problem statement, all variables and given/known data

    24 = 10(ex/20 + e-x/20)

    2. Relevant equations



    3. The attempt at a solution

    2.4 = ex/20 + e-x/20

    ln2.4 ≠ x/20 - x/20

    I tried using ex/20 = ex - e20 but I don't think this is right either. Came up with

    20.69 - ln e-x = x

    and this is just another silly answer. I must be forgetting some rule. It seems to me like the variables keep canceling out, which I know can't be right.
     
  2. jcsd
  3. May 24, 2012 #2
    Do you know the representations for the hyperbolic functions?
     
  4. May 24, 2012 #3
    I guess not.
     
  5. May 24, 2012 #4

    Mark44

    Staff: Mentor

    The step above is incorrect. There is NO property of logs that allows you to simplify log(A + B) to logA + logB.

    Multiply all three terms in the original equation by ex/20. That will give you an equation that is quadratic in form. With a suitable substitution, you can use the Quadratic Formula to solve that equation.
     
  6. May 24, 2012 #5
    To expand on Mark's idea, if ex/20 = u, then what does the equation change into?
     
  7. May 24, 2012 #6
    Thanks Mark! I had tried multiplying by e^x/20 and using the quadratic forumula but I guess I didn't do it right the first time. Checked the trash can: I made a calculator error and didnt notice it. Hate it when I do something right and then make a little mistake that makes me think im not doing it right and I end up trying things that I know look wrong. Thanks again.
     
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