A bones age ( a math question)

Question

The radioactive element​ carbon-14 has a​ half-life of 5750 years. A scientist determined that the bones from a mastodon had lost

83.5​%

of their​ carbon-14. How old were the bones at the time they were​ discovered?

Answer 1

0.165 = (0.5)^t/5750

ln 0.165 = (t / 5750) ln 0.5

-1.80 ≈ (t / 5750) (-0.70)

2.57 ≈ t / 5750

14.78 ≈ t

Answer 2

Carbon₁₄ half-life 5,750 yrs. Bones lost 83.5% of C_14,, so still retain 16.5% of C₁₄.

N_t = N₀/(2^t/hl) N_t = Amt of C₁₄ after t yrs; N₀ = original amt of C₁₄. hl = half-life = 5,750 yrs.

2^t/5750= N₀/ N_t

N_t/N₀ = 0.165 so N₀/N_t = 1/0.165 = 6.0606

2^t/5750 = 6.0606 -> take ln of both sides: t/5750 • ln(2) = ln(6.0606)

t/5750(0.6931) = 1.8018 Solve for t: t = 1.8018(5750/0.6931) ≈ 14,947 yrs. (Depends on h.m. decimal places used in the calc'n).