To the unknown base

Let's evaluate the following wigglish expression :

${\displaystyle 3_x{4}_x{2}_{x}5.}$

Telescoping from the left, we get

${\displaystyle 3_x{4}_x{2}_{x}5 = {(3x + 4)}_x{2}_{x}5,}$
${\displaystyle = {(3x^2 + 4x + 2)}_{x}5,}$
${\displaystyle = 3x^3 + 4x^2 + 2x + 5.}$

So in brief, we get

${\displaystyle 3_x{4}_x{2}_{x}5 = 3x^3 + 4x^2 + 2x + 5.}$

The pattern that suggests itself here is true in general---a polynomial in the variable ${\displaystyle x}$ can be thought of as a number in base ${\displaystyle x}$.

Polynomials can be added, subtracted, and multiplied as if they were numbers in base ${\displaystyle x}$.  For instance

${\displaystyle (3x^3 + 4x^2 + x + 5) + (x^3 + 2x^2 + 7x +2)}$
${\displaystyle = 4x^3 + 6x^2 + 8x + 7}$

becomes

${\displaystyle 3_x{4}_x{1}_{x}5 + 1_x{2}_x{7}_{x}2}$
${\displaystyle = 4_x{6}_x{8}_{x}7.}$

The last calculation resembles the ordinary addition ${\displaystyle 3415+1272 = 4687.}$

The product

${\displaystyle (x+2)(x+3) = x^2+5x+6}$

can be rendered

${\displaystyle \left(1_{x}2\right)\left(1_{x}3\right) = 1_x{5}_{x}6}$

and this resembles the ordinary numerical product ${\displaystyle 12\times 13 = 156.}$

Base ${\displaystyle x}$ calculations do not always agree with corresponding decimal calculations.  The base ${\displaystyle x}$ calculation is most often actually simpler, because there is no carrying, as there with ordinary numbers, e.g. :

${\displaystyle \left(15\right)\left(15\right) = 1^1{0}^{2}5 = 225,}$

but, (no carrying steps this time),

${\displaystyle \left(1_{x}5\right)\left(1_{x}5\right) = 1_x{10}_{x}25.}$