Solving for 'a' in a Complex Equation

Fe%5E4%29+%2B+%28a%2Fe%5E-4%29+%2B+a+%3D+7In summary, the given equation, 7 = a/2(e^4/a + e^-4/a) + a, does not have a real valued solution because when graphed, there is no intersection between the equation and the line y=7.
  • #1
Denyven
19
0

Homework Statement



Solve for a

[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

Homework Equations



[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

The Attempt at a Solution



[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

multiply by 2 on both sides

[tex]14=a(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

this is where i am stuck

is it even possible?
 
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  • #2
There certainly won't be a nice closed formula for a and there may be no real value of a that solves it at all.
 
  • #3
Denyven said:

Homework Statement



Solve for a

[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

Homework Equations



[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

The Attempt at a Solution



[tex]7=\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

multiply by 2 on both sides

[tex]14=a(e^\frac{4}{a} + e^\frac{-4}{a}) + a[/tex]

this is where i am stuck

is it even possible?

when you multiplied with 2, you didnt multiply the last term, which should give 2a
 
  • #4
hint: don't multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..
 
  • #5
thebigstar25 said:
hint: don't multiply with 2, and use your knowledge of Hyperbolic functions.. things may get easier from there ..
That's a good suggestion but I doubt things will get much easier!

[tex]\frac{e^{\frac{4}{a}}+ e^{-\frac{4}{a}}}{2}= cosh(\frac{4}{a})[/tex]
but there is still a problem with that "a" outside the cosh.
 
  • #6
Hmmm Alright. So I now have
[tex]7=a cosh(\frac{4}{a})+a[/tex]
I guess I would move a over and divide by a to get
[tex]\frac{7-a}{a}=cosh(\frac{4}{a})[/tex]
Now I don't know what to do. Is there some property of cosh in which I can pull something out or split something? We just started hyperbolic functions and I still know very little.
Thanks very much for all the help so far!
 
  • #7
There is no real valued solution. Here's a graph of the relevant portion of

[tex]
\frac{a}{2}(e^\frac{4}{a} + e^\frac{-4}{a}) + a\hbox{ and } 7
[/tex]



graph.jpg
 

Related to Solving for 'a' in a Complex Equation

1. What is the process for solving for 'a' in a complex equation?

The process for solving for 'a' in a complex equation involves isolating the variable 'a' on one side of the equation and simplifying the other side using algebraic techniques such as distributing, combining like terms, and using inverse operations.

2. How do I know when to use the quadratic formula to solve for 'a'?

The quadratic formula is used to solve for 'a' when the equation is in the form of ax2 + bx + c = 0. It is also used when other methods, such as factoring or completing the square, cannot be used to solve the equation.

3. Can I use a calculator to solve for 'a' in a complex equation?

Yes, calculators can be used to solve for 'a' in a complex equation, but it is important to understand the steps involved in solving the equation by hand in order to check the accuracy of the calculator's answer.

4. What are the common mistakes to avoid when solving for 'a' in a complex equation?

Common mistakes to avoid when solving for 'a' in a complex equation include forgetting to perform the same operation on both sides of the equation, making a sign error when simplifying, and forgetting to apply the order of operations.

5. Can I use the quadratic formula to solve for 'a' in any complex equation?

No, the quadratic formula can only be used to solve for 'a' when the equation is in the form of ax2 + bx + c = 0. Other methods, such as factoring or completing the square, must be used for equations in different forms.

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