The rock's age is approximately #"2.6 billion"# years.
There are essentially two ways of solving nuclear half-life 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...
