Answer: 2500 years
Explanation:
I'm not quite sure if I'm doing this right myself but I'll give it a shot.
We use this formula to find half-life but we can just plug in the numbers we know to find t.
![A(t)=A_(0)(1/2)^t^/^h](https://img.qammunity.org/2020/formulas/mathematics/high-school/mkufy3imzcf9ne2gi8rfjbkyhorlrghxp0.png)
We know half-life is 5730 years and that the parchment has retained 74% of its Carbon-14. For
let's just assume that there are 100 original atoms of Carbon-14 and for A(t) let's assume there are 74 Carbon-14 atoms AFTER the amount of time has passed. That way, 74% of the C-14 still remains as 74/100 is 74%. Not quite sure how to explain it but I hope you get it. h is the last variable we need to know and it's just the half-life, which has been given to us already, 5730 years, so now we have this.
![74=100(1/2)^t^/^5^7^3^0](https://img.qammunity.org/2020/formulas/mathematics/high-school/ylevuli3eq8rshyoukezj0csaebfpnql8e.png)
Now, solve. First, divide by 100.
![0.74=(0.5)^t^/^5^7^3^0](https://img.qammunity.org/2020/formulas/mathematics/high-school/n084b9gexebu7b27hnro9c3w8uztusm3ml.png)
Take the log of everything
![log(0.74)=(t)/(5730) log(0.5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/c659wil2ftn5a1r8f2qdgy7elafulgycou.png)
Divide the entire equation by log (0.5) and multiply the entire equation by 5730 to isolate the t and get
![5730(log(0.74))/(log(0.5)) =t](https://img.qammunity.org/2020/formulas/mathematics/high-school/unps4wlic4zz4zwtwp7222nccu120bzl55.png)
Use your calculator to solve that giant mess for t and you'll get that t is roughly 2489.128182 years. Round that to the nearest hundred years, and you'll find the hopefully correct answer is 2500 years.
Really hope that all the equations that I wrote came out good and that that's right, this is definitely the longest answer I've ever written.