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pderugin

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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

pderugin

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770 posts
Joined 11/2009

Also I should add that in the back of the book, there were three solutions: 0.85, 0.96, 9.92

Posted about 1 year ago

Befeltingu

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209 posts
Joined 12/2009

What class is this for? Can you do it graphically? I dont know how to do i algebraically but if you just graph e^(x) = x^3(2x + 1) and see where they intercept.

Posted about 1 year ago

pderugin

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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 Undecided

Posted about 1 year ago

huntse

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1432 posts
Joined 11/2010

1)I would think that any cubic equation has (up to) 3 solutions, no?
2)by "solve" they mean algebraically. You can check with a graph but they're not going to give you much credit.

Posted about 1 year ago

TecmoSuperBowl

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I got to X = log(2x^4 + x^3) before I realized I don't really remember what I'm doing.

Posted about 1 year ago

Befeltingu

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209 posts
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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

identifier

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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

Befeltingu

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209 posts
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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.



I do love the shelf that contains all my old textbooks

Posted about 1 year ago

huntse

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1432 posts
Joined 11/2010

I dont think that this problem is solvable algebraically especially if you cant use calc.



Well then fuck it, just get out newton raphson. (spot the person with the engineering mindset)

Posted about 1 year ago

EUSSI

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1990 posts
Joined 06/2010

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

PrinzVonHapunkt

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1198 posts
Joined 12/2010

dont you get 2x^4 + X^3 if you multiply the right side with X^3?

If you then ln both sides you get

X = ln(2) + 7ln(X) if I'm right

but I dunno how to proceed

Posted about 1 year ago

BaseMetal

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2051 posts
Joined 01/2010

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

huntse

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1432 posts
Joined 11/2010

Perhaps that's the point of the question. To show you that you need to learn calculus.

Posted about 1 year ago

shuttle

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3333 posts
Joined 11/2008

I was thinking a taylor series approach might be the go here, but tbh I'm not sure there's an algebraic solution here.

Posted about 1 year ago




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