What more do you want? There is no algebraic method to solve mx= sin x.

In general, there is no algebraic method to solve x= f(x) where f is any transcendental function.

Results 1 to 8 of 8

- November 1st 2009, 12:03 PM #1

- Joined
- Apr 2009
- Posts
- 190

- November 2nd 2009, 02:32 AM #2

- Joined
- Apr 2005
- Posts
- 17,994
- Thanks
- 2355

- November 2nd 2009, 02:37 AM #3

- Joined
- Apr 2009
- Posts
- 190

That's all the question said, so I am guessing that we are just meant to consider that there are 3 solutions if the gradient is less than 1, or else there will be just 1. I wasn't sure what else we could do with it.

There was also another question on the sheet which was similar that I thought couldn't be solved:

**Solve a^b=b^a for all real a and b**

but it might be solvable through logs?

- November 2nd 2009, 05:49 AM #4

- November 2nd 2009, 05:50 AM #5
Hello AquafinaAre you sure about that? See the attached diagram.

There was also another question on the sheet which was similar that I thought couldn't be solved:

**Solve ab=ba for all real a and b**

but it might be solvable through logs?

Grandad

- November 2nd 2009, 06:16 AM #6

- Joined
- Apr 2009
- Posts
- 190

- November 2nd 2009, 07:00 AM #7

- November 2nd 2009, 07:24 AM #8