...wait, how? I’ve never heard of a use like this.

 If f is monotonic and continuous, you can solve f(x) = y for x by successively narrowing an interval [a, b] where f(a) <= y <= f(b) (or vice versa for decreasing functions). In the case that the range of f spans the entire real numbers, the initial range can be determined as the smallest range [-2^k, 2^k] where k is an integer (a strategy often used for open-ended binary search).
 Interestingly, with floats you can also binary search on the binary representation. That narrows down the answer to the most precise answer representable in floating point in at most 64 steps for 64 bit floats. If you bisect the interval in the middle at each step, then it could take more steps (possibly more than 1000).It turns out that doing this is pretty much equivalent to what you suggest, if you narrow down the initial range [-2^k, 2^k] by binary searching on k, because that's just binary searching on the exponent in the binary representation.
 Interesting! I didn't really think about this but this is how I implemented sqrt() and division in an interview. Didn't think of how it generalizes!

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