Evaluating (ln^2) in Easy Math Problem: X=Xo*e^(ln^2/Td)*T | Answer: Xo=17

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So if

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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Wait ##\ln ^2##? That's not defined. Maybe it should be just ln(2)
 
I used ln(2) but that did not give me an answer of 17
 
Nvm I got. Thanks
 
I think it would have funny if you had posted some unproven conjecture and then moments later posted "nvm, got it."

Just a thought..
 
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