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 5 - 2011 - Paper 3

From the information given, when the leaf is 4 days old, its mass is 2.5 grams. We can substitute $m = 2.5$ and $t = 4$ into the equation:

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

Dividing both sides by 7.5 gives:

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

Taking the natural logarithm of both sides results in:

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

This implies:

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

Thus, we find:

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

To find $t$, we start by deriving the equation for the rate of change of mass:

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

Substituting $k$ and $p$ gives:

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

Setting this equal to $-0.6 \ln 3$ leads to:

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

Both sides can be divided by $-\ln 3$:

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

Solving for $e^{-\left( \frac{1}{4} \ln 3 \right)t}$ gives:

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

Taking the natural logarithm of both sides results in:

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

which simplifies to:

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

Calculating the numerical approximation provides:

$t \approx 4.146...$

Thus, $t$ is approximately 4.15.