Solving a Tricky Calculus Equation: 24 = 10(ex/20 + e-x/20)

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The discussion revolves around solving the calculus equation 24 = 10(ex/20 + e-x/20). The original poster struggles with the problem, mistakenly deriving ln2.4 = x - x and x = √(400ln2.4) ≈ 18.7, while knowing the correct answer is x ≈ 12.5. A suggestion is made to multiply all terms by ex/20 to transform the equation into a quadratic form, allowing the use of the Quadratic Formula for a solution. The poster acknowledges a previous calculator error that led to confusion and incorrect assumptions about their process. Overall, the conversation emphasizes the importance of careful calculations and proper application of algebraic techniques.
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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.

Homework Statement



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

Homework Equations





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.
 
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Do you know the representations for the hyperbolic functions?
 
I guess not.
 
Larsani said:
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.

Homework Statement



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

Homework Equations





The Attempt at a Solution



2.4 = ex/20 + e-x/20

ln2.4 ≠ x/20 - x/20
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.
Larsani said:
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.
 
To expand on Mark's idea, if ex/20 = u, then what does the equation change into?
 
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 I am not doing it right and I end up trying things that I know look wrong. Thanks again.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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