exponential equation

[sxandr6]

e^x+e^-x=5

 

When I first tried to solve this I tried to take the natural log but that ends up cancelling the x's out. Then I tried to square both sides but I think I messed up squaring the left half of the equation. I ended up with: e^2x+e^x+2e=25. Thoughts?

[ajb]

[math]Log[a+b] \neq Log[a] + Log[/math]

 

If you used that.

 

Also, recall that

 

[math]Cosh[x] = \frac{1}{2}(e^{x} + e^{-x})[/math]

 

and that

 

[math]ArCosh[x] = Log[x + \sqrt{x^{2}-1}][/math]

Edited November 10, 2009 by ajb

[hermanntrude]

EDIT: the question is answered above. It's better than my answer by a long way

[sxandr6]

no, I tried: Ln e^x + Ln e^-x= Ln 5 and then the Ln e = 1

so there for x-x= Ln 5. But I don't think that's right.

[ajb]

no, I tried: Ln e^x + Ln e^-x= Ln 5 and then the Ln e = 1

so there for x-x= Ln 5. But I don't think that's right.

 

That is not right as log is not a linear function.

[sxandr6]

Yeah, I didn't think that was right. But the explanation above is over my head. I'm only in college algebra. I thought squaring both sides would work because the back of the book give 2 answers: -1.567 and 1.567.

[ajb]

The most direct way is to use the arcosh. Have you discussed hyperbolic functions before? If not what about the circular functions?

[sxandr6]

Nope, sorry not that far either. I might just ask the teacher when I see him next. It's not due for a little while.

[ajb]

You could solve the problem graphically.

 

Plot

 

[math]y = \frac{1}{2}(e^{x} + e^{-x})[/math]

 

and you get

 

You can now examine what values of x for which y = 5/2.

 

In particular notice that the plot is symmetric about x=0. Thus the two values you quote.

[sxandr6]

Ok I graphed it and found the answer, but where did the 1/2 come from. Is it a property of the cosh.

[the tree]

That's just how cosh is defined.

[ajb]

As the tree says.

[Bignose]

Most of the time when hyperbolic trig functions have not been introduced yet, the intention of the book is to make you think of good substitutions to use. In this case, I suspect that you were supposed to let [math]u=e^x[/math].

 

Then you get a quadratic in u, which you can solve, and then as the final step re-translate the u back into x.