We are given that the half-life of radioactive carbon $^{14}C$ is 5730 years. A parchment has 74% of the $^{14}C$ radioactivity compared to plant material today. We need to determine the age of the parchment.

1. Problem Description

We are given that the half-life of radioactive carbon $^{14}C$ is 5730 years. A parchment has 74% of the $^{14}C$ radioactivity compared to plant material today. We need to determine the age of the parchment.

2. Solution Steps

The formula for radioactive decay is given by:

$N(t) = N_0 e^{-kt}$, where $N(t)$ is the amount of the radioactive substance at time $t$, $N_0$ is the initial amount of the substance, and $k$ is the decay constant.

The half-life $T_{1/2}$ is the time it takes for the amount of the substance to reduce to half of its initial value. Thus, $N(T_{1/2}) = \frac{1}{2} N_0$.

Substituting this into the decay formula, we get:

$\frac{1}{2} N_0 = N_0 e^{-kT_{1/2}}$

$\frac{1}{2} = e^{-kT_{1/2}}$

Taking the natural logarithm of both sides:

$ln(\frac{1}{2}) = -kT_{1/2}$

$k = \frac{ln(2)}{T_{1/2}}$

In this problem, $T_{1/2} = 5730$ years. Therefore, $k = \frac{ln(2)}{5730}$.

The parchment has 74% of the $^{14}C$ radioactivity as plant material does today. This means that $N(t) = 0.74 N_0$.

We want to find the age of the parchment, i.e., the time $t$.

$0.74 N_0 = N_0 e^{-kt}$

$0.74 = e^{-kt}$

Taking the natural logarithm of both sides:

$ln(0.74) = -kt$

$t = \frac{-ln(0.74)}{k}$

Substituting $k = \frac{ln(2)}{5730}$:

$t = \frac{-ln(0.74)}{\frac{ln(2)}{5730}} = \frac{-ln(0.74) \cdot 5730}{ln(2)}$

$t = \frac{-(-0.3011)}{0.6931} \cdot 5730 = \frac{0.3011}{0.6931} \cdot 5730$

$t \approx 0.4344 \cdot 5730 \approx 2489.412$

Therefore, the age of the parchment is approximately 2489 years.

3. Final Answer

The age of the parchment is approximately 2489 years.