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Equation:X=Xo*[e^(ln^2)/Td] * T

with

X = 5e12

Td = 15

T= 672

And the answer is Xo=17

My question:is how do I evaluate (ln^2) ??

Usually when I use the ln function to calculate something its like ln(5/2)^2 or something like that but here its just ln^2

what does this mean? How do you get 17 for Xo

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# Easy math problem

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