The mass, m grams, of a leaf t days after it has been picked from a tree is given by $$m = pe^{-kt}$$ where k and p are positive constants - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 3

Given that the mass after 4 days is 2.5 grams:

$2.5 = 7.5 e^{-4k}$

Dividing both sides by 7.5:

$\frac{2.5}{7.5} = e^{-4k}$ $\frac{1}{3} = e^{-4k}$

Taking the natural logarithm:

$-4k = \ln(\frac{1}{3})$ $k = -\frac{1}{4} \ln(3)$

Using the formula for the rate of change of mass:

$\frac{dm}{dt} = -kpe^{-kt}$

Substituting k and p:

$-0.6 \ln 3 = -\left(\frac{1}{4} \ln 3\right)(7.5)e^{-\left(\frac{1}{4} \ln 3\right)t}$

Rearranging gives:

$\frac{0.6 \ln 3}{\frac{1}{4} \ln 3 \times 7.5} = e^{-\left(\frac{1}{4} \ln 3\right)t}$

Simplifying:

$\frac{0.6 \cdot 4}{7.5} = e^{-\left(\frac{1}{4} \ln 3\right)t}$

Calculating the left side:

$\frac{2.4}{7.5} = e^{-\left(\frac{1}{4} \ln 3\right)t}$

Taking the logarithm of both sides:

$-\left(\frac{1}{4} \ln 3\right)t = \ln\left(\frac{2.4}{7.5}\right)$

Thus,

$t = -\frac{4}{\ln 3} \ln\left(\frac{2.4}{7.5}\right)$

Calculating gives approximately:

$t \approx 4.146...$