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## Help me with this math problem

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#### 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.

#### pderugin

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

#### Befeltingu

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.

#### 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

#### huntse

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.

#### TecmoSuperBowl

5546 posts
Joined 01/2009

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

#### Befeltingu

209 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

#### 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.

#### Befeltingu

209 posts
Joined 12/2009

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

#### huntse

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)

#### EUSSI

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

#### PrinzVonHapunkt

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

#### BaseMetal

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.

#### huntse

1432 posts
Joined 11/2010

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

#### shuttle

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.

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