Question

Let \(x=x(t)\) be the solution curve of the differential equation \[ \frac{dx}{dt}=-kx, \] with \[ x(0)=100,\quad x\!\left(\frac{1}{2}\right)=80. \] If \(x(t_\alpha)=5\), then \(t_\alpha\) is equal to:

• A. \(\displaystyle \frac{\ln 5+\ln 4}{2(\ln 5-\ln 4)}\)
• B. \(\displaystyle \frac{\ln 5+\ln 4}{\ln 5-\ln 4}\)
• C. \(\displaystyle \frac{\ln 5-\ln 4}{2(\ln 5+\ln 4)}\)
• D. \(\displaystyle \frac{\ln 5-\ln 4}{\ln 5+\ln 4}\)

Hint:

For exponential decay equations \(\frac{dx}{dt}=-kx\), use ratios of given values to eliminate the constant \(C\) quickly and find \(k\).

Answer 1

The Correct Option is A

Step 1: Solve the differential equation \[ \frac{dx}{dt}=-kx \Rightarrow \frac{dx}{x}=-k\,dt \] \[ \ln x=-kt+C \Rightarrow x=Ce^{-kt} \] Step 2: Use initial condition \(x(0)=100\) \[ 100=C \Rightarrow x(t)=100e^{-kt} \] Step 3: Use the second condition \[ x\!\left(\frac12\right)=80 \Rightarrow 100e^{-k/2}=80 \] \[ e^{-k/2}=\frac{4}{5} \Rightarrow -\frac{k}{2}=\ln\frac{4}{5} \Rightarrow k=2(\ln5-\ln4) \] Step 4: Find \(t_\alpha\) when \(x(t_\alpha)=5\) \[ 100e^{-kt_\alpha}=5 \Rightarrow e^{-kt_\alpha}=\frac{1}{20} \] \[ -kt_\alpha=\ln\frac{1}{20}=-(\ln4+\ln5) \] \[ t_\alpha=\frac{\ln4+\ln5}{k} \] Substitute \(k=2(\ln5-\ln4)\): \[ t_\alpha=\frac{\ln5+\ln4}{2(\ln5-\ln4)} \] \[ \boxed{\displaystyle \frac{\ln5+\ln4}{2(\ln5-\ln4)}} \]