Question

-0.0001211 Use the model (t)-20e for radiocarbon dating to answer the question. A sample from a mummified bull was taken from a certain place. The sample shows that 86% of the carbon-14 still remains. How old is the sample? Round to the nearest year. The sample is approximately years old.

Answer 1

Let the sample in staring Q_o = 100 unit

after years it remain Q(t) = 86 units

$Q(t)=Q_oe^{-0.000121t}$

86 = 100 e -0.000121t

switch sides

$100\cdot e^{-0.000121t}=86$

divide 100 both sides

$\frac{100}{100}\cdot e^{-0.000121t}=\frac{86}{100}$

$e^{-0.000121t}=0.86$

Apply log both sides

$log(e^{-0.000121t})=log(0.86)$

Apply log rule log_xa^b = b.log_xa

$(-0.000121\cdot t)log\;e=log(0.86)$

Apply log_ee = 1

$-0.000121\cdot t=log(0.86)$

divide -0.000121 both sides

$t=\frac{log(0.86)}{-0.000121}$

$t=\frac{-0.15082288}{-0.000121}$

The sample is approximately 1246 years old.