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E to the y plus ...trial and error math? 
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#1
Feb1814, 01:41 PM

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A co worker has the following equation:
[itex]3e^{.5y} +3e^{y} + 3e^{1.5y} + 103e^{2y} =98.39[/itex] Solve for y. Some sort of compound interest bond equation I am told, or something like that. He has been told that to solve for y, one must use a trial and error approach. True?? 


#2
Feb1814, 02:15 PM

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If you substitute u for [itex] e^{ \frac {1}{2} y} [/itex] you get a quartic equation, for wich an exact solution exists. Type the equation in WolframAlpha to get a meaningless jumble of really large numbes and lots of square and cube root signs.
It's less work to solve the quartic with a numerical method like newton's method than to use the formula for the quartic equation. 


#3
Feb1814, 02:17 PM

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There is an exact solution. Let ##x=\exp(y/2)##. Then your equation is equivalent to ##103x^4+3x^3_3x^2_3x98.39=0##. This is a quartic equation, so it is solvable, exactly. Then solve for y. Simple!
Not so simple. Solving cubics is a bear of a problem. Solving quartics? That's a megafauna bear of a problem. Solving this numerically is nontrivial. Newton's method doesn't work very well on this problem. You need to use something else such as Edit: The secant method doesn't work very well here either because f(x) is almost flat between 1 and +1. 


#4
Feb1814, 03:04 PM

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E to the y plus ...trial and error math?
willem2 and DH...thanks! This forum is loaded with some very brilliant minds. He couldn't find an answer anywhere online or from his college finance professors, so I told him not to worry, I would get an answer through the best site on the web.
Thanks again! 


#5
Feb1814, 03:07 PM

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Let ##x=\exp(y/2)##. Then your equation is equivalent to ##103x^4+3x^3+3x^2+3x98.39=0##. 


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Feb1814, 03:12 PM

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Feb1814, 04:04 PM

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#8
Feb1814, 05:15 PM

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