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.

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- Nov 1st 2009, 01:03 PM #1

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- Nov 2nd 2009, 03:32 AM #2

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- Nov 2nd 2009, 03:37 AM #3

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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?

- Nov 2nd 2009, 06:49 AM #4

- Nov 2nd 2009, 06:50 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

- Nov 2nd 2009, 07:16 AM #6

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- Nov 2nd 2009, 08:00 AM #7

- Nov 2nd 2009, 08:24 AM #8