Solving 24 cosh x + 16 sinh x = 2500

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To solve the equation 24 cosh x + 16 sinh x = 2500, one can use the identities for cosh x and sinh x in terms of e^x. By substituting these identities, the equation simplifies to 20e^x + 4e^(-x) = 2500. This leads to a quadratic equation in the form 20y^2 - 2500y + 4 = 0 after substituting y = e^x. Solving this quadratic yields two potential solutions for y, but only y = 125 results in a valid real number for x. Therefore, the solution to the original equation is x = ln(125).
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I am having trouble solving for x

24 cosh x + 16 sinh x = 2500

do I multiply out all the (e)s?
 
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That's one option:
Following that, multiply the equation with e^{x} ;
this gives you a quadratic equation in the variable e^{x}
 
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To solve this equation, you can use the identity cosh x = (e^x + e^(-x))/2 and sinh x = (e^x - e^(-x))/2. This will give you:

24((e^x + e^(-x))/2) + 16((e^x - e^(-x))/2) = 2500

Simplifying this equation, you will get:

12e^x + 12e^(-x) + 8e^x - 8e^(-x) = 2500

Combining like terms, you will get:

20e^x + 4e^(-x) = 2500

Now, you can substitute y = e^x and solve for y:

20y + 4/y = 2500

Multiplying both sides by y, you get a quadratic equation:

20y^2 + 4 = 2500y

Subtracting 2500y from both sides, you get:

20y^2 - 2500y + 4 = 0

Solving this quadratic equation, you will get two solutions: y = 0.002 or y = 125.

Substituting back y = e^x, you will get two solutions for x: x = ln(0.002) or x = ln(125).

Since ln(0.002) is not a real number, the only solution for x is x = ln(125).

So, the solution to the equation 24 cosh x + 16 sinh x = 2500 is x = ln(125).
 
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