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

[bmed90]

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

 

[pwsnafu]

Wait $\ln ^2$? That's not defined. Maybe it should be just ln(2)

 

[bmed90]

I used ln(2) but that did not give me an answer of 17

 

[bmed90]

Nvm I got. Thanks

 

[DeeAytch]

I think it would have funny if you had posted some unproven conjecture and then moments later posted "nvm, got it."

Just a thought..