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

From the mass function:

At $t = 0$, $m = 7.5$ grams,

At $t = 4$, $m = 2.5$ grams:

2.5 = 7.5 e^{-4k} \ \frac{1}{3} = e^{-4k} \ -4k = \text{ln}\left(\frac{1}{3}\right) \ k = -\frac{1}{4} \text{ln}(3)$$

Using the given mass function, we differentiate:

$\frac{dm}{dt} = -k p e^{-kt}$

Substituting for $k$ and $p$:

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

Equating to $-0.6 \text{ln} 3$ gives:

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

Solving for $t$:

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

Letting $x = e^{\frac{1}{4} \text{ln}(3)t}$, we find:

t \approx 4.146 \text{or around } 4.1$$