Find the smallest $b$ for that $b^x - \log_b x$ has exactly one zero I tried to find the smallest $1.3 < b < 1.5$ for that $f(x) = b^x - \log_b x$ has only one zero (root) for $x > 0$, but all I managed to do is:
$f'(x) = \ln (b) \cdot b^x - \frac{1}{x \ln (b)}$
Now, setting this equal to zero should give the tip of the function-plot. When by variation of $b$ this is exactly on the x-axis this should be the value of $b$.
However, I don't know how to proceed after this step.
EDIT: I maybe found a start. This could be made more strict.
Since $0 = b^x - \log_b x$, then $b^x = \log_b x$.
The right side is just the inverse function of the left side. They are equal, and the only way this is possible is that $b^x = x$, because that is what determines the inverse function.
Lets make this into $g(x) = b^x - x$ as a function of $x$.
$g'(x) = \ln (b) b^x - 1$
This can be transformed to
$b^x = \frac{1}{\ln (b)}$
and
$x = \log_b (\frac{1}{\ln (b)})$
This is the value of $x$ for that $g(x)$ is the smallest. Substituting the x in the equation in the beginning, this yields:
$0 = b^{\log_b (\frac{1}{\ln (b)})} - \log_b (\frac{1}{\ln (b)})$
Which gets us to
$b^{\frac{1}{\ln (b)}} = \frac{1}{\ln (b)}$
Now, if we solve this for $b$, this should be the desired result. The left side is for some weird reason always equal to $e$, but I don't know why.
EDIT2:
Say that $b^{\frac{1}{\ln (b)}} = c$. We now show that $c = e$.
$b^{\frac{1}{\ln (b)}} = c$
$b = c^{\ln (b)}$
$\log_c (b) = \ln (b)$
This shows that $b^{\frac{1}{\ln (b)}} = e$.
That gets us to
$e = \frac{1}{\ln (b)}$
and
$b = e^{\frac{1}{e}}$
which is what we want.
 A: Only a single root
means that it is tangent.
Therefore both
$f(x) =0$
and
$f'(x) = 0$.
Since
$f'(x) 
= \ln (b) \cdot b^x - \frac{1}{x \ln (b)}
$,
so that
$\ln (b)  b^x
=\frac{1}{x \ln (b)}
$
or
$\ln^2 (b)  b^x
=\frac{1}{x}
$.
Since $f(x) = 0$,
$b^x = \frac{\ln(x)}{\ln(b)}
$,
we have
$\ln^2 (b) \frac{\ln(x)}{\ln(b)}
=\frac{1}{x}
$
or
$\ln(b)
=\frac1{ x \ln(x)}
$.
I see that
$x=e, b=e^{1/e}$
works,
but showing this is the
only solution
is annoyingly hard.
Staring at this hasn't helped,
so I'll leave it at this
and hope that 
someone else
can see what I have missed.
A: Nice. So, is this a fair re-cap?

We seek $b$ and $x$ for which the following both vanish
  $$
\begin{aligned}
f(x) &= b^x - log_b(x) \\
f'(x) &= b^x \, \ln (b)   - \frac{1}{x \ln (b)}
\end{aligned}
$$

Setting $f(x)=0$ gives  $b^x = log_b(x)$. We have $b>0$ so $b^{x}$ is well behaved for $x \in \mathbb{R}$ and $log_{b}(x)$ is well defined for $x>0$.  
Denote $g(x) = b^x$.
So $g^{-1}(x) = log_{b}(x) 
\, \Rightarrow
g(x) = g^{-1}(x)
\, \Rightarrow 
g(x) = x
\, \Rightarrow 
b^{x} = x
$
Substitute $b^{x} = x$ into $f'(x) = 0$ to get 
$$
\begin{aligned}
x \, \ln (b)  &- \frac{1}{x \ln (b)}  = 0 \\
x \, \ln (b)  &= \frac{1}{x \ln (b)} \\
x \, \ln (b) &= \pm 1 \\
e^{x \, \ln (b)} &= e^{\pm 1} \\
b^{x } &= x = e^{\pm 1}
\end{aligned}
$$
Choosing $x=e^1$ gives 
$$
e = b^{e} \Rightarrow 1 = e \ln(b) \Rightarrow \ln(b) = 1/e \Rightarrow b = e^{1/e}
$$
Choosing $x=e^{-1}$ gives 
$$
e^{-1} = b^{e^{-1}} \Rightarrow -1 = e^{-1} \ln(b) \Rightarrow \ln(b) = -e \Rightarrow b = e^{-e}
$$
So there seem to be two sets of solutions for $(b,x)$ and the question asks for the first set. 
A: A plot of some iso-curves of $b^x - log_b(x) = z$ for various $z$ values below. The curve we are interested in is $b^x - log_b(x) = 0$. The line $b = e^{1/e}$ is shown in red and touches this curve at its maximum. This is the only value of $b$ for which there is one solution in $x$ along the curve (at $x=e$). I can find this value using a numeric method but would be very interested to see an analytic approach.

