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Posted by on Monday, January 19, 2009 at 4:17am.

how do i solve for t:

-.50=1-e^-(t/125)^5

i got this far:

ln(.5)=5ln(-t/125)

t=?

  • math ln - , Monday, January 19, 2009 at 5:59am

    5 e^-(t/5^2) = 1.5
    e^-(t/125) = .3
    e^(t/125) = 1/.3 = 3.3333
    t/125 = ln 3.3333
    t = 150.5

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