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The value of a dollar decreases by 23% every year. Find the half-life in this situation. The half-life is years. (Do not roun
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\\ $Let $y(t) $ be the value of dollar at time $ t. \\\\ $We are given that the value of dollar decreases by $23\%$ every year$\\\\ $Hence, the rate is $r = 0.23\\\\ $Hence, we have$\\\\ \frac{dy}{dt} = -0.23y \\\\ \Rightarrow \frac{dy}{y} = -0.23 \;dt\\\\ \\ \Rightarrow \int \frac{dy}{y} = -\int0.23 \;dt\\\\\\ \Rightarrow \ln |y| = -0.23\cdot t + C \;\;\;\;\text{ where }C $ is integration constant$ \\\\ $Since $y\geq 0 \rightarrow \ln |y| = \ln y\\\\ \Rightarrow \ln y = -0.23\cdot t + C\\\\ \Rightarrow y = e^{-0.23\cdot t + C} = e^C e^{-0.23\cdot t}\\\\ $Let, $e^C = K \\\\ \Rightarrow y(t) = K e^{-0.23\cdot t} \;\;\;\ \ldots(1)\\\\ $Let, the initial value of dollar be $Y_0 \\\\ \Rightarrow y(0) = Y_0\\\\ $Putting $ t= 0 $ and $ y = Y_0 $ in (1) we get$

\\\\ Y_0 = K e^{-0.23\cdot 0}= K \\\\ \Rightarrow K = Y_0 \;\;\; \ldots(2)\\\\ $Substituting (2) into (1) we get$\\\\ y(t) = Y_0 e^{-0.23\cdot t}\;\;\;\ldots(3) \\\\ $Let the half life time be $\tau \\\\ \Rightarrow y(\tau) = \frac{y(0)}2 \\\\ \Rightarrow y(\tau)= \frac{Y_0}{2}\\\\ $Substituting $t= \tau $ and $y=\frac{Y_0}{2} $ in (3) we get$\\\\ \frac{Y_0}{2} = Y_0 e^{-0.23\cdot \tau}\\\\ \Rightarrow \frac12 = e^{-0.23\cdot \tau}\\\\ \Rightarrow e^{0.23\cdot \tau} = 2\\\\ \Rightarrow 0.23\cdot \tau = \ln 2\\\\ \Rightarrow \tau = \frac{\ln 2}{0.23} \simeq 3.014 \simeq 3 \;\;\;\;$(Rounded to nearest tenth)$\\\\ \mathbf{The \;required \;answer \;you\; need \;to \;enter \;is:\;\;3 }

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