Suppose you find a rock originally made of potassium-40, half of which decays into argon-40 every 1.25 billion years. You open the rock and find 7 atoms of argon-40 for every atom of potassium-40. How long ago did the rock form?

3.75 billion years.

Explanation:

Let the amount of potassium-40 at the beginning be K.

We can follow the decay with its half life.

In 1 half life, the amount of potassium-40 left is K/2

In 2 half lives, the amount left will be K/2²

In 3 half lives, it will be K/2³

In n half lives, it will be K/2ⁿ

So, in n half lives, the amount of Argon left will be (K - (K/2ⁿ))

In the question, it is described that the amount of Argon left = 7 × the amount of potassium-40 left

(K - (K/2ⁿ)) = 7 (K/2ⁿ)

K(1 - 2⁻ⁿ) = 7K/2ⁿ

1 - 2⁻ⁿ = 7(2⁻ⁿ)

Divide through by 2⁻ⁿ

(1/2⁻ⁿ) - 1 = 7

2ⁿ = 7+1 = 8

2ⁿ = 2³

n = 3

This means, we'll get 7 atoms of argon-40 for every atom of potassium-40 in 3 half life

1 half life = 1.25 billion years

3 half lives = 3 × 1.25 billion years = 3.75 billion years.