Finding out the permissible value of $k$

Question

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.

Answer 1

Let us rewrite

$$k=a^{1/a}=b^{1/b},$$ and for convenience

$$\frac1k=\left(\frac1a\right)^{1/a}=\left(\frac1b\right)^{1/b}.$$

So we are looking for the number of roots of the equation

$$x^x=\frac1k.$$

By canceling the derivative,

$$(\ln x+1)x^x=0,$$ we have a signle extremum at $x=e^{-1}$ so that $x^x=e^{-1/e}$. The function starts from $x=0,x^x=1$, goes through the extremum then increases to infinity, so there can only be two solutions when

$$k\in[1,e^{1/e}].$$

Answer 2

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

$$f(x_0)>0$$

$$f''(x_0)>0$$