The mass of an isotope decreases at a rate that is proportional to the mass at that time. $\newline$ The mass of the isotope was $40$ grams initially, and it was $10$ grams after $16$ days. $\newline$ What was the mass of the isotope after $20$ days? $\newline$ Round to the nearest gram. $\newline$ $\square$ grams

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Exponential growth and decay: word problems

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Q. The mass of an isotope decreases at a rate that is proportional to the mass at that time.

\newline

The mass of the isotope was

40

grams initially, and it was

10

grams after

16

days.

\newline

What was the mass of the isotope after

20

days?

\newline

Round to the nearest gram.

\newline

\square

grams

Identify Mass and Time: Identify the initial mass, the mass after a certain time, and the time elapsed. $\newline$ Initial mass $m_0$ = $40$ grams $\newline$ Mass after $16$ days $m_{16}$ = $10$ grams $\newline$ Time elapsed for $m_{16}$ = $16$ days $\newline$ We need to find the mass after $20$ days $m_{20}$ .
Use Exponential Decay Formula: Use the formula for exponential decay, which is $m(t) = m_0 \cdot e^{-kt}$ , where $m(t)$ is the mass at time $t$ , $m_0$ is the initial mass, $k$ is the decay constant, and $t$ is the time. $\newline$ We need to find the value of $k$ using the information that $m(16) = 10$ grams.
Substitute Values and Solve: Substitute the known values into the exponential decay formula to find $k$ . $10 = 40 \cdot e^{-16k}$ Divide both sides by $40$ to isolate $e^{-16k}$ . $\frac{10}{40} = e^{-16k}$ $\frac{1}{4} = e^{-16k}$
Take Natural Logarithm: Take the natural logarithm of both sides to solve for $k$ . $\ln(\frac{1}{4}) = \ln(e^{-16k})$ $\ln(\frac{1}{4}) = -16k \times \ln(e)$ Since $\ln(e) = 1$ , we have: $\ln(\frac{1}{4}) = -16k$
Solve for k: Solve for k. $\newline$ $k = -\ln(\frac{1}{4}) / 16$ $\newline$ $k \approx 0.08664$ (using a calculator)
Find Mass After $20$ Days: Now that we have $k$ , we can find the mass after $20$ days using the same exponential decay formula. $m(20) = 40 \cdot e^{(-0.08664 \cdot 20)}$ Calculate the value using a calculator. $m(20) \approx 40 \cdot e^{(-1.7328)}$ $m(20) \approx 40 \cdot 0.177$ $m(20) \approx 7.08$
Round to Nearest Gram: Round the result to the nearest gram. $m(20) \approx 7$ grams