Determine the number base $x$ if $(59.5)_{(x)} = (89.3125)_{(10)}.$

Question

How can I determine the base number system $x$ of two decimal numbers using an equation?

Answer 1

Recall that a place-value numeral is just a polynomial (or a Laurent polynomial if there are digits after the point) at the base. Thus, "59.5" in base $x$ means $5x + 9 + 5x^{-1}$. So you just need to solve:

$$5x + 9 + 5x^{-1} = 89.3125$$ $$5x^2 + 9x + 5 = 89.3125x$$ $$5x^2 - 80.3125x + 5 = 0$$

Or, multiplying by $\frac{16}{5}$ so that everything is an integer

$$16x^2 - 257x + 16 = 0$$ $$x = \frac{257 \pm \sqrt{(-257)^2 - 4(16)(16)}}{2(16)}$$ $$x = \frac{257 \pm \sqrt{65025}}{32}$$ $$x = \frac{257 \pm 255}{32}$$ $$x \in \lbrace \frac{1}{16}, 16 \rbrace$$

Although fractional bases are a thing, they're way less common than integer bases, so 16 is probably the intended answer.

Answer 2

Well, a freebie... as $5 < 8$ we know that $x > 10$.

But basically we have $5x + 9 + \frac 5x = 89.3125$

Since $\frac 5x < 1$ we have the $5x +9$ must be an integer and $5x+9 = 89$ and so $x =16$.

That's all. That's the only thing it could be.

But that's not enough. We have to verify that $x=16$ is possible. After all, just because we asked a question doesn't mean there is actually an answer.

So we need to verify that $\frac 5{16} = 0.3125$ which ... it does.