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The rate of decay of a radioactive substance is proportional to the amount present. In $1840$ there were $50$ grams of the substance and in $1910$ there were $35$ grams. How many grams of the substance remain in $1990$ ?

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Write and solve direct variation equations

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Q. The rate of decay of a radioactive substance is proportional to the amount present. In

1840

there were

50

grams of the substance and in

1910

there were

35

grams. How many grams of the substance remain in

1990

?

Understand and Determine Type of Variation: Understand the problem and determine the type of variation. The problem states that the rate of decay of a radioactive substance is proportional to the amount present. This is an example of exponential decay, which can be represented by the equation $A = A_0 \cdot e^{(-kt)}$ , where $A$ is the amount of substance at time $t$ , $A_0$ is the initial amount of substance, $e$ is the base of the natural logarithm, $k$ is the decay constant, and $t$ is the time elapsed.
Find Decay Constant: Use the given information to find the decay constant $k$ . We know that in $1840$ , there were $50$ grams of the substance, and in $1910$ , there were $35$ grams. This gives us two points in time: $t_0 = 0$ (taking $1840$ as the starting point) with $A_0 = 50$ grams, and $t_1 = 1910 - 1840 = 70$ years with $A_1 = 35$ grams. We can use these values to find $k$ .
Set Up Equation to Solve: Set up the equation with the known values to solve for $k$ . Using the exponential decay formula: $35 = 50 \cdot e^{(-k \cdot 70)}$ .
Solve for k: Solve for k. $\newline$ Divide both sides by $50$ : $\frac{35}{50} = e^{(-k \cdot 70)}$ . $\newline$ Simplify the fraction: $0.7 = e^{(-k \cdot 70)}$ . $\newline$ Take the natural logarithm of both sides: $\ln(0.7) = \ln(e^{(-k \cdot 70)})$ . $\newline$ Use the property of logarithms: $\ln(0.7) = -k \cdot 70 \cdot \ln(e)$ . $\newline$ Since $\ln(e) = 1$ , we have: $\ln(0.7) = -70k$ . $\newline$ Divide by $-70$ : $k = \frac{-\ln(0.7)}{70}$ . $\newline$ Calculate k: $k \approx \frac{-(-0.356675)}{70} \approx 0.005095$ .
Find Remaining Amount in $1990$ : Use the value of $k$ to find the amount of substance remaining in $1990$ . $\newline$ The time from $1840$ to $1990$ is $t = 1990 - 1840 = 150$ years. We can now use the decay constant $k$ and the initial amount $A_0 = 50$ grams to find the amount $A$ remaining in $1990$ .
Substitute Values and Calculate: Substitute the values into the exponential decay formula. $\newline$ $A = 50 \cdot e^{-0.005095 \cdot 150}$ . $\newline$ Calculate the amount: $A \approx 50 \cdot e^{-0.76425}$ . $\newline$ Calculate $e^{-0.76425}$ : $e^{-0.76425} \approx 0.46575$ . $\newline$ Multiply by $50$ : $A \approx 50 \cdot 0.46575$ . $\newline$ Calculate $A$ : $A \approx 23.2875$ .

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