pderugin
770 posts
Joined 11/2009
I just can't figure it out ... I keep going around in circles. If someone could be so kind as to explain it to me and/or show me the steps, that would be awesome!
Solve the equation:
(e^x) / (x^3) = 2x + 1
State each solution correct to two decimal places.
Posted about 1 year ago
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pderugin
770 posts
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Befeltingu
205 posts
Joined 12/2009
pderugin
770 posts
Joined 11/2009
It's for a precalc class. The question didn't specify how they want it solved. I'll try graphing it, thank you.
EDIT: The graphs of the two functions, e^(x) and x^3(2x + 1), confirm two of the three answers, 0.85, 0.96. Just wish it weren't so difficult algebraically 
Posted about 1 year ago
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huntse
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TecmoSuperBowl
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Befeltingu
205 posts
Joined 12/2009
I dont think that this problem is solvable algebraically especially if you cant use calc. (i.e. you could use power series expantion to approximate e^x = 1 + x + x^2/2 ....). I did graph it though and did find the 3 solutions you listed. Try zooming out on your graph because the 3rd solution is way the F out there.
Ya i just dont know how to do this algebraically sorry. You could PM bellatrix haha she probably has lots of free time
Posted about 1 year ago
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identifier
2141 posts
Joined 07/2008
I got to X = log(2x^4 + x^3) before I realized I don't really remember what I'm doing.
I got to 0=2x^4+x^3 - e^x, realized I was doing it wrong then decided to not bother getting the huge textbook down from the shelf.
Posted about 1 year ago
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Befeltingu
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huntse
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EUSSI
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I just can't figure it out ... I keep going around in circles. If someone could be so kind as to explain it to me and/or show me the steps, that would be awesome!
Solve the equation:
(e^x) / (x^3) = 2x + 1
State each solution correct to two decimal places.
not sure (its been a while)
but if you multiply both sides by x^3 & add -1 you get :
e^x - 1 = 2x^4
lets say e^x - 1 = t
then x = log(1+1/t)
then x = log (1+1/2x^4)
and now you can go further i think
btw i could be completely wrong
Posted about 1 year ago
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PrinzVonHapunkt
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BaseMetal
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I am not certain but I think this equation is a Transcendental Equation, ie, a pure algebraic approach won't work. To find the roots you could try an iterative approach like Newton-Raphson but then this involves guessing start points and calculating the tangents at some points and for this you would use calculus, so it's a bit odd for a pre-calc question. (The roots are -0.85, +0.92, +9.92).
The graphical approach looks pretty reasonable except it's tricky getting to 2 decimal places.
Posted about 1 year ago
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huntse
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shuttle
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