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