The equationhas a solution somewhere between 1 and 2, since if
and if
If we call this solutionthensatisfies
We can rearrange this equation in various ways. Two such are
and
We can use these rearrangements as iteration formulae to attempt to findto 2 decimal places say. The iteration formulae are
and
Withthese formulae give respectively
Iterate 

1.5 
1.5 

1.5753 
1.447 

1.683 
1.409 

1.839 
1.383 

2.071 
1.364 

2.436 
1.351 

3.070 
1.342 

4.421 
1.335 

9.011 
1.330 

90.485 
1.327 
The progress of the iterates is shown below in the two cases. In the graph below left, the gradient of the graph is greater than 1 and the graph crosses the linefrom below. The iterates diverge. In the graph below right the gradient of the graph is less than 1 and the graph crosses the line from above. The iterates converge to