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The "x-figure" notation is essentially logarithmic with an offset: "6-figures" = 10^5. Therefore, "mid-six-figures" should be roughly 10^5.5 or $316227.
by bmm6o 16y ago
The "x-figure" notation is essentially logarithmic with an offset: "6-figures" = 10^5. Therefore, "mid-six-figures" should be roughly 10^5.5 or $316227.
- simcop2387 16y agoI believe that's the geometric mean he's referring to. (a*b)^1/2 # geometric mean, square root of the product of the two numbers (10^5 * 10^6)^(1/2) # substitute in for 10^5 and 10^6, the limits of the six-figure range (10^(5+6))^(1/2) # simplify inner product (10^11)^(1/2) # more simplify 10^(11 * 1/2) # move the 1/2 in, and ... 10^(11/2) # simplify 10^5.5 # and finally you get the same result The geometric mean always struct me as an interesting idea in mathematics