The radioactive substance Iridium-192 has a half-life of 73.83 days. How many days would it take for a sample of Iridium-192 to decay to 10% of its original amount? Round your answer to the nearest integer.
10% =100% * (0.5)^(d/73.83), solve for d 0.1 =0.5^(d/73.83) Take the log of both sides log(0.1) =d/73.83 * log(0.5) d/73.83 =log(0.1) / log(0.5) d/73.83 =3.3219281..... d =73.83 * 3.3219281... d =~245 - days for given amount to decay to 10%.
10% =100% * (0.5)^(d/73.83), solve for d 0.1 =0.5^(d/73.83) Take the log of both sides log(0.1) =d/73.83 * log(0.5) d/73.83 =log(0.1) / log(0.5) d/73.83 =3.3219281..... d =73.83 * 3.3219281... d =~245 - days for given amount to decay to 10%.
