The mass, m grams, of a leaf t days after it has been picked from a tree is given by $$m = p e^{-kt}$$ where k and p are positive constants - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 3

Given that the mass after 4 days is 2.5 grams, we can substitute into the formula:

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

Dividing both sides by 7.5:

rac{1}{3} = e^{-4k}

Taking the natural logarithm on both sides:

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

Therefore,

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

The rate of change of mass with respect to time is:

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

Substituting the values of p and k:

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

Simplifying gives:

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

Thus,

$e^{-\left( \frac{1}{4} \ln 3 \right)t} = \frac{0.6 \times 4}{7.5} = \frac{2.4}{7.5} = 0.32$

Taking the natural logarithm:

$-\left( \frac{1}{4} \ln 3 \right)t = \ln(0.32)$

From which:

$t = -\frac{4 \ln(0.32)}{\ln 3}\approx 4.146... \text{or } 4.15.$