I was given the equation;
$(x-7)^a=1$
where $a=(x-4)$
The 3 solutions are: $x=4, 6, 8$
When $x=4$,
$(-3)^0=1$, which can be reached by setting $(x-4)=0$ because $n^0=1$
When $x=8$,
$1^4=1$, which can be reached by setting $(x-7)=1$ because $1^n=1$
The less obvious one and the one I was not able to reach algebraically but rather just noticed was $x=6$, which produces;
$(-1)^2=1$, because $(-1)^n=1$ such that n is an even integer
How could I reach $x=6$ algebraically?
Well since we have $1$ at the RHS, then we have three possible ways achieve it using exponentials. First is to set the exponent to $0$, so we have:
$$x-4 = 0 \implies x=4$$
Second is to set the base to $1$, since $1^k = 1; \forall k$
$$x-7 = 1 \implies x=8$$
The third is to set the base to $-1$, since $(-1)^{2n} = 1; \forall n$
$$x-7 = -1 \implies x=6$$
Now since $6$ is an odd number, we get that it's a solution.