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It depends on the precise nature of the coder and there are many possibilities. Here are three:
A: f(x) = 2x
B: f(x) = 4*(x - 1)
C: f(x) = 2*(x2 - 3x + 4)
The simplest of these is f(x) = 2x.
In that case the decoder is y = ln(x)/ln(2) where ln is the natural logarithm. Logs to base 10, or any other base can also be used and, if the chosen base is 2, then the decoder becomes:
y = log2(x)
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∙ 11y ago
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Is 4 and 2 thirds - 2 thirds greater than 4 eighths plus 3 and 2 eighths?
4 and 2/3 - 2/3 = 4 4/8 + 3 and 2/8 is the same as 3 + 2/4 +1/4 which is 3 and 3/4 4 > 3 + 3/4 (aka, yes)
What is -2 wholes 3 over 4 minus 1 over 4?
-2 3/4 - 1/4 = -(2 3/4 + 1/4) = -[2 (3+1)/4] = -(2 4/4) = -(2 + 1) = -3 or -2 3/4 - 1/4 = - (4*2 + 3)/4 - 1/4 = -11/4 - 1/4 = (-11 - 1)/4 = -12/4 = -3
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