The rate of decay of a certain substance is d | vedclass

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(D) Let $x$ be the amount of the substance left at time $t$. The rate of decay is $\frac{dx}{dt} = -kx$,where $k > 0$. Integrating the differential eq

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$\frac{16}{3} 	ext{ gms}$

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(D) Let $x$ be the amount of the substance left at time $t$. The rate of decay is $\frac{dx}{dt} = -kx$,where $k > 0$. Integrating the differential equation: $\int \frac{1}{x} dx = \int -k dt \implies \ln x = -kt + C$. At $t = 0$,$x = 27$,so $C = \ln 27$. Thus,$\ln x = -kt + \ln 27$,or $\ln(\frac{x}{27}) = -kt$. At $t = 3$,$x = 8$,so $\ln(\frac{8}{27}) = -3k$. Since $\frac{8}{27} = (\frac{2}{3})^3$,we have $\ln((\frac{2}{3})^3) = -3k \implies 3 \ln(\frac{2}{3}) = -3k \implies k = -\ln(\frac{2}{3}) = \ln(\frac{3}{2})$. Substituting $k$ back: $\ln(\frac{x}{27}) = -t \ln(\frac{3}{2}) = t \ln(\frac{2}{3})$. For $t = 4$,$\ln(\frac{x}{27}) = 4 \ln(\frac{2}{3}) = \ln((\frac{2}{3})^4) = \ln(\frac{16}{81})$. Therefore,$\frac{x}{27} = \frac{16}{81} \implies x = 27 	imes \frac{16}{81} = \frac{16}{3} 	ext{ gms}$.

The rate of decay of a certain substance is directly proportional to the amount present at that instant. Initially,there are $27 \text{ gms}$ of the substance and $3 \text{ hours}$ later it is found that $8 \text{ gms}$ are left. The amount left after one more hour is:

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