Select branch: 3

The equation is: lnx+(k/x)-k+2 = 0 (x >; 1)

Let f(x)=lnx+(k/x)-k+2.

f'(x)=( 1/x)-(k/x？ )=(x-k)/x？

x∈( 1，k)，f '(x)& lt； 0, on which simply subtract f (x), the range (f(k), 2).

The function of x∈(k, +∞), f'(x) >; 0, f(x) increases on it, and the range of values is (f(k), +∞).

f'(k)=0，f(k)=lnk-k+3

The necessary and sufficient condition for the original equation to have a unique real root is:

lnk-k+3=0

Let g (k) = lnk-k+3 (k >; 3)

g '(k)=( 1/k)- 1 & lt； 0, g(k) is simply reduced on (3, +∞).

And g (4) = ln (4/e) >: 0, g(5)=ln(5/e? )& lt0

The root of the equation lnk-k+3=0 is in (4,5).

So the value of k is within (4,5).