Question

seth was studying a population of fruit flies. the population, p, after t days of the study is modeled by the function p(t)=10e^{0.5t}. what function could seth use to find d, the day the number of fruit flies reached a given value, p?
a. d = \frac{1}{10}\ln(2p)
b. d = \frac{1}{5}\ln(p)
c. d = \frac{1}{2}\ln(10p)
d. d = 2\ln(\frac{p}{10})
e. d = 10\ln(\frac{p}{2})

Explanation:

Step1: Set $P(t) = p$

We have $p = 10e^{0.5t}$.

Step2: Isolate the exponential term

Divide both sides by 10: $\frac{p}{10}=e^{0.5t}$.

Step3: Apply natural - logarithm to both sides

$\ln(\frac{p}{10})=\ln(e^{0.5t})$. Since $\ln(e^{x}) = x$, we get $\ln(\frac{p}{10}) = 0.5t$.

Step4: Solve for $t$ (which is $D$)

$t = D=\frac{\ln(\frac{p}{10})}{0.5}$. Since $0.5=\frac{1}{2}$, then $D = 2\ln(\frac{p}{10})$.

Answer:

D. $D = 2\ln(\frac{p}{10})$