$1011 \textrm{(base }t) = 4931 \textrm{(base 10)}$
I have to find $t$, which is the base of 1011.
I do the following:
$4931 \textrm{(base 10)} = 4 \times 10^3 + 9 \times 10^2 + 3 \times 10^1 + 1 \times 10^0$
$= 4000 + 900 + 30 + 1$
$= 4931$
$1011 \textrm{(base }t) = 1 \times t^3 + 0 \times t^2 + 1 \times t^1 + 1 \times t^0$
$= t^3 + 0 + t + 1$
$= t^3 + 1 + 1$
$t^3 + t + 1 = 4931$
$t^3 + t = 4930$
Is there a mistake made here? $t$ should equal the base but I have multiple solutions in this case.
I used a method similar to what is in this thread: Finding which base number given operations
The polynomial $t^3 + t -4930 = 0$ has only one real solution, $t=17$.