Can ln(u)=u Be Solved for x Algebraically?

  • Thread starter Thread starter danielatha4
  • Start date Start date
AI Thread Summary
The equation ln(u) = u, where u is a function of x, cannot be solved for x algebraically. This conclusion is based on the nature of logarithmic and exponential functions. While numerical methods or graphical solutions may provide approximate values, an exact algebraic solution is not feasible. The discussion emphasizes the complexity of the relationship between logarithmic and linear functions. Therefore, algebraic manipulation does not yield a solution for x in this case.
danielatha4
Messages
113
Reaction score
0

Homework Statement



Can this equation be solved for x? This isn't any type of homework. I'm doing this for fun. This equation came from an integration while solving a differential equation.
 

Attachments

  • danseeqn.jpg
    danseeqn.jpg
    13.2 KB · Views: 498
Physics news on Phys.org
I'm afraid not. In general, ln(u)=u where u is a function of x cannot be solved for x algebraically.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Replies
14
Views
2K
Replies
6
Views
2K
Replies
21
Views
3K
Replies
7
Views
3K
Replies
10
Views
2K
Back
Top