The 'Unlock All Digits' Game

Solution 1:

No. Since you have as many operations as digits, you must use each operation once and only once, which means all operands must also be digits. But the only numbers expressible as roots (of the form $\sqrt[a]{b}$ (for $a\gt 1$, otherwise nothing new is gained) are $2$ and $3$, and likewise the only numbers expressible as logs are $0$, $1$, $2$ and $3$. Now, from just $\{0\}$ the only number that's accessible is $1$ and from $\{0,1\}$ the only number that's accessible is $2$; and you can't get to $1$ from $\{0\}$ using $\log$, and you can't get to $2$ from $\{0,1\}$ using $\sqrt{}$ or $\log$. This leaves you without enough operations left over for the rest of the digits.

Note that it's possible to complete the chain starting from other digits. For instance, from $9$ one can go $\log_9(9)=1$, $S(1)=2$, $\sqrt[2]{9}=3$, $2^3=8$, $\frac82=4$, $4+1=5$, $2\cdot 3=6$, $P(8)=7$, $7-7=0$.