Digvijay is right, $f(x)$ will be minimum at $x = 2,$
Here is another approach.
Since log and exponent are monotonically increasing functions, the problem of minimizing $f(x)$ can be reduced to just minimizing the quadratic expression
$x^{2} - 4x + 5,$
this quadratic expression can be written as $(x^{2} - 4x + 4) + 1$ which is equal to $(x - 2)^{2} + 1.$
now since $(x - 2)^{2}$ can not be less than $0$, so $(x - 2)^{2} + 1$ can not be less than $1.$
Also $(x - 2)^{2} + 1$ will be at its minimum value $(= 1)$, when $x = 2.$
so value of $f(x)$ will be minimum at $x = 2.$