The rock’s age is approximately ##”2.6 billion”## years.

There are essentially two ways of solving problems. One way is by applying the half-life formula, which is

##A(t) = A_0(t) * (1/2)^(t/t_(1/2))## , where

##A(t)## – the quantity that remains and has not yet decayed after a time t; ##A_0(t)## – the initial quantity of the substance that will decay; ##t_(1/2)## – the half-life of the decaying quantity;

In this case, the rock contains ##”1/4th”## of the orignal amount of potassium-40, which means ##A(t)## will be equal to ##(A_0(t))/4##. Plug this into the equation above and you’ll get

##(A_0(t))/4 = A_0(t) * (1/2)^(t/t_(1/2))##, or ##1/4 = (1/2)^(t/t_(1/2))##

This means that ##t/t_(1/2) = 2##, since ##1/4 = (1/2)^2##.

Therefore,

##t = 2 * t_(1/2) = 2 * “1.3 = 2.6 billion years”##

A quicker way to solve this problem is by recognizing that the initial amount of the substance you have is halved with the passing of each half-life, or ##t_(1/2)##.

This means that you’ll get

##A = (A_0)/2## after the first 1.3 billion years

##A = (A_0)/4## after another 1.3 billion years, or ##2 * “1.3 billion”##

##A = (A_0)/8## after another 1.3 billion years, or ##2 * (2 * “1.3 billion”)##

and so on…