sxandr6 Posted November 10, 2009 Share Posted November 10, 2009 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? Link to comment Share on other sites More sharing options...
ajb Posted November 10, 2009 Share Posted November 10, 2009 (edited) [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 Link to comment Share on other sites More sharing options...
hermanntrude Posted November 10, 2009 Share Posted November 10, 2009 EDIT: the question is answered above. It's better than my answer by a long way Link to comment Share on other sites More sharing options...
sxandr6 Posted November 10, 2009 Author Share Posted November 10, 2009 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. Link to comment Share on other sites More sharing options...
ajb Posted November 10, 2009 Share Posted November 10, 2009 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. Link to comment Share on other sites More sharing options...
sxandr6 Posted November 10, 2009 Author Share Posted November 10, 2009 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. Link to comment Share on other sites More sharing options...
ajb Posted November 10, 2009 Share Posted November 10, 2009 The most direct way is to use the arcosh. Have you discussed hyperbolic functions before? If not what about the circular functions? Link to comment Share on other sites More sharing options...
sxandr6 Posted November 10, 2009 Author Share Posted November 10, 2009 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. Link to comment Share on other sites More sharing options...
ajb Posted November 10, 2009 Share Posted November 10, 2009 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. Link to comment Share on other sites More sharing options...
sxandr6 Posted November 10, 2009 Author Share Posted November 10, 2009 Ok I graphed it and found the answer, but where did the 1/2 come from. Is it a property of the cosh. Link to comment Share on other sites More sharing options...
the tree Posted November 10, 2009 Share Posted November 10, 2009 That's just how cosh is defined. Link to comment Share on other sites More sharing options...
ajb Posted November 10, 2009 Share Posted November 10, 2009 As the tree says. Link to comment Share on other sites More sharing options...
Bignose Posted November 11, 2009 Share Posted November 11, 2009 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. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now