Geologic Time Help! EQUATION IS: T = (1/λ) * (ln (Astandard/Asample))

Algebra
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Sep 9th, 2015

Hi there! Thank you for the opportunity to help you with your question!

T = 1/0.0001244 * ln(1/6.25) = 8033 * ln(0.16) = 8033*(-1.833) = -14,724yrs

T = 1/0.0001244 * ln(.5/6.25) = 8033 * ln(0.08) = 8033*(-2.526) = -20,291yrs

T = 1/0.0001244 * ln(.25/6.25) = 8033 * ln(0.04) = 8033*(-3.219) = -25,858yrs

T = 1/0.0001244 * ln(.125/6.25) = 8033 * ln(0.02) = 8033*(-3.912) = -31,425yrs

T = 1/0.0001244 * ln(.05/6.25) = 8033 * ln(0.008) = 8033*(-4.828) = -38,783yrs

T = 1/0.0001244 * ln(.001/6.25) = 8033 * ln(0.00016) = 8033*(-8.74) = -70,208yrs

T = 1/0.0001244 * ln(.0005/6.25) = 8033 * ln(0.00008) = 8033*(-9.433) = -75,775yrs

T = 1/0.0001244 * ln(.00004/6.25) = 8033 * ln(0.00000064) = 8033*(-11.959) = -96066yrs


Please let me know if you need any clarification. Always glad to help!
Sep 9th, 2015

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