Why does this problem have one less root for x?

Question

Problem statement (Problem $6$):

This is my work below:

The answer of $x$ is $2$ and $8$. but my answer is $1$/ $\sqrt{3}$ and it is a credible answer if you solve it. Why is this happening?

Answer 1

When $x=1/\sqrt3$, $x-1<0$ and so $(x-1)^7<0$. But $|x-1|>0$ and so $|x-1|^{\textrm{nasty expression}}>0$. Thus $1/\sqrt3$ cannot be a solution.

Answer 2

Remember that you are looking for a solution under the condition $(x-1)>0$.

Note that since $t=\log_3 x$ the solutions are $x=81$ and $x=\frac1{\sqrt3}$.

As pointed out by quasi, the other solution come from the trivial case $x-1=1$.

Answer 3

There are two answers, $81$ and $2$.

If $a$ is positive, and $s,t$ are real, the equation $$a^s = a^t$$ holds if and only if $a=1$ or $s=t$.

You analyzed the case $s=t$, which yields $x=81$.

You missed the case $a=1$, which yields $x=2$.