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 5 - 2011 - Paper 3

Given the mass equation:

$m = p e^{-kt}$

Substituting in the values:

When $t = 0$, $m = 7.5$:

$7.5 = p e^{-k(0)} = p$

Then, when $t = 4$, $m = 2.5$:

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

Dividing both sides by 7.5 gives:

e^{-4k} = rac{2.5}{7.5} = rac{1}{3}

Taking the natural logarithm:

-4k = ext{ln } rac{1}{3}

This implies:

k = -rac{1}{4} ext{ln } 3

Thus, we can conclude:

k = rac{1}{4} ext{ln } 3

We start with the rate of change of mass given by:

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

Substituting $k = \frac{1}{4} \text{ln } 3$ and $p = 7.5$ gives:

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

Setting this equal to (-0.6 \text{ln } 3$$:

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

Cancelling the negative signs and dividing both sides by $\text{ln } 3$ gives:

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

Solving for $e^{-\frac{1}{4} \text{ln } 3 t}$ leads to:

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

Calculating:

$= \frac{0.6}{1.875} = 0.32$

Taking the natural logarithm of both sides:

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

This implies:

$\text{ln } 3 t = -4 \cdot \text{ln }(0.32)$

Finally, solving for $t$ gives:

$t = \frac{-4 \cdot \text{ln }(0.32)}{\text{ln } 3}$

Calculating the specific values will yield:

$t \approx 4.146...$ or $t \approx 4.15$.