$y=\sqrt{x}$ and $y=\ln(x)$ have no points of intersection for $x>0$, while $y=\sqrt[3]{x}$ and $y=\ln(x)$ have $2$ such intersection points with $x>0$.
$y=\sqrt[p]{x}$ and $y=\ln(x)$ have exactly one point of intersection (for $x>0$).
Find $p$ and $x_0$.
On
My instinct would be reverse the variables $x$ and $y$, so that you are looking at the inverse functions $y=x^p$ and $y=e^x$. You are seeking intersection points $x^p=e^x$.
This is equivalent to finding the zeros of $f(x)=e^x-x^p$. Evidently there are two zeros whenever this continuous function successfully dips below the $x$-axis, giving a solution on the way down and another on the way up.
There is a clear location where $p=e$ where you can show that $e^x-x^e\ge0$ and equality only when $x=e$.
Because I swapped $y$ and $x$, the answer to the original problem occurs when $p=e$ and $y_0=e$ and $x_0=e^e$.
Your question shouldn't have solution. Let me explain: for every $x\in[0,1]$, $x^{1/p}>\log(x)$. Moreover
$$ \lim_{x\to\infty} \frac{\log(x)}{\sqrt[p]x} = \lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{p}\frac{1}{x^{(p-1)/p}}} = \lim_{x\to\infty}p \frac{x^{(p-1)/p}}{x} =\lim_{x\to\infty} p \frac{1}{\sqrt[p] x} = 0 . $$
This results means that, given $p$, there exists $M>0$ such that
$$ \sqrt[p] x >\log (x) \qquad \forall x>M . \tag{$\ast$} $$
So we have: for $0<x<1$ $$ \sqrt[p] x >\log (x) $$ and for $x>M$ $$ \sqrt[p] x >\log (x) . $$
If exists a point $01x_0<M$ such that
$$ \sqrt[p]{x_0} = \log(x_0) $$ and that $$ \sqrt[p]{x} = \log(x) $$ for $x\in(x_0,x_0+\epsilon)$ with $\epsilon>0$ enough small, then it must exist $x_0<x_0'<M$ such that $$ \sqrt[p]{x} = \log(x) $$ and $$ \sqrt[p]{x} >\log(x) $$ for $x\in(x_0',x_0'+\epsilon')$, otherwise equation$(*)$ doens't hold. So intersection points appear in pairs.
However, I can't proof there is no point in which both functions have the same tangent line, and I'm not sure how to disprove that.