Finding out the permissible value of $k$ 
If $\log_ka=a$ and $\log_kb=b$ for exactly two distinct positive real numbers $a$ and $b$,then $k$ can't be equal to
$A)e^{\frac1{2e}}$
$B)e^{\frac2e}$
$C)e^{\frac12}$
$D)e^{\frac13}$

Let $f(x)=\log_ex\log_ke-x$ .We are looking for only two solutions $a$ and $b$ of this equation.For this to happen $f'(x)$ must vanish at one point between $a$ and $b$ and $f''(x)$ at this corresponding point should not be zero otherwise it would indicate a point of inflexion.I tried differentiating but couldnot see how do I find out the plausible range of values of $k$.Any ideas?Thanks.

We may rewrite as follows:
$$f(x)=k^x-x$$
And we want to show there is only one or no root for some $k$ by finding the extrema i.e. $f'(x_0)=0$.
$$f'(x)=\ln(k)k^x-1=0\implies x_0=\frac{\ln\left(\frac1{\ln(k)}\right)}{\ln(k)}$$
And show that for one of those given values of $k$ that
$$f(x_0)>0$$
$$f''(x_0)>0$$
