Potassium-40 has a half-life of 1.25 billion years (by) and decays to argon-40. how old is a fossil that has a 40k/40ar ratio of ~ 26%?

You can do a very quick estimation by telling the ~ 26% is close to 25%, end then two half-life have passed: one from 100% to 50% concentration, and other from 50% to 25% concentration. So, 2 * 1.25 billion years = 2.50 billion years.

The answer, then is that the fossil is 2.50 by old.

Given that this method has an accuracy of + / - 10% this answer is good enough.

For didactical purposes, I am goint to show you the exact procedure.

C = Co * e^ ( - kt)

Half-life time = > C = Co / 2

=> C / Co = (1/2) = e ^ (-kt)

=> - kt = ln (1/2) = > kt = ln (2)

t = 1.25 by = > k (1.25) = ln (2) = > k = ln (2) / 1.25 = 0.5545

=> C/Co = e ^ (-kt)

In the problem C/Co = 26/100 = > 0.26 = e^ (-0.5545t)

=> - 0.5545t = ln (0.26)

=> t = - ln (0.26) / 0.5545 = 2.43 by.

So with the exact procedure you obtain 2.43 by while with the estimation that ~26% is close to 25% you obtain 2.50 by.