While traveling to Europe, phelan exchanged 250 US dollars

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sana20

19.02.2022

Mathematics

The half-life of carbon-14 is 5730 years.decay equation: for initial value $a_0$ , the amount of carbon-14 left after $t$ years is $a = a_0\left(\dfrac{1}{2}\right)^{t/5730}$ we are given that $a = 0.27a_0$ (final amount is 27%, which is 0.27, of the intial)so $\begin{aligned} 0.27a_0 & = a_0\left(\dfrac{1}{2}\right)^{t/5730} \end{aligned}$ divide both sides by $a_0$ $\begin{aligned} 0.27 & = \left(\dfrac{1}{2}\right)^{t/5730} \end{aligned}$ take log base 10 of both sides $\begin{aligned} \log(0.27) & = \log\left[\left(\dfrac{1}{2}\right)^{t/5730} \right] \end{aligned}$ use logarithm property $\log m^n = n \log m$ $\begin{aligned} \log(0.27) & = \log\left[\left(\dfrac{1}{2}\right)^{t/5730} \right] \\ \log(0.27) & = \frac{t}{5730} \log(0.5) \\ \log(0.27) & = t \cdot \frac{\log(0.5)}{5730} \\ t & = \frac{5730 \log(0.27)}{\log(0.5)} \\ t & \approx 10823.79\text{ years} \end{aligned}$ so approximately 10824 years old.

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