Prove that for `lambda gt 1`, the equation `xlog x +x =lambda` has least one solution in `[1 , lambda]`.

Explanation

Consider `f(x)=(x-lambda)logx" for "1lexle lambda`
The function f is continuous on `[1,lambda]`
and `f'(x)=logx+(x-lambda)/(x)" for all "x in (1, lambda)`
Also, `f(1)=0 and f(lambda)=0`.
So, by Rolle's theorem, there is `c in (1, lambda)` such that `f'(c)=0`.
`rArr" "logc+(c-lambda)/(c)=0`
`rArr" "clog c+(c -lambda)=0`
`rArr" "clogc+c=lambda`
`rArr" "c log c+c=lambda`
Hence, `log x+x=lambda` has at least one solution in `[1,lambda]`.