Cerium-144 is a radioactive isotope with a half-life of 285 days. How long would it take for an initial sample of 15 grams of Cerium-144 to decay until only 5 gram remains?a. about 452 days b. about 855 days c. about 570 days d. about 407 days

To solve this we are going to use the formula: [tex]A=A_{0}( \frac{1}{2} )^{ \frac{t}{h} [/tex]
where 
[tex]A[/tex] is the final amount aster a time [tex]t[/tex]
[tex]A_{0}[/tex] is the initial amount 
[tex]t[/tex] is the time 
[tex]h[/tex] is the half-life

We know for our problem that [tex]A=5[/tex], [tex]A_{0}=15[/tex], and  [tex]n=285[/tex], so lets replace those values in our formula:
[tex]A=A_{0}( \frac{1}{2} )^{ \frac{t}{h} [/tex]
[tex]5=15( \frac{1}{2} )^{ \frac{t}{285} [/tex]

Since [tex]t[/tex] is the exponent, we are going to use logarithms to find its value:
[tex]5=15( \frac{1}{2} )^{ \frac{t}{285} [/tex]
[tex]( \frac{1}{2} )^{ \frac{t}{285}}= \frac{5}{15} [/tex]
[tex]( \frac{1}{2} )^{ \frac{t}{285}}= \frac{1}{3} [/tex]
[tex]ln( \frac{1}{2} )^{ \frac{t}{285}}= ln(\frac{1}{3})[/tex]
[tex] \frac{t}{285} ln( \frac{1}{2})=ln( \frac{1}{3} )[/tex]
[tex] \frac{t}{285}= \frac{ln( \frac{1}{3}) }{ln( \frac{1}{2} )} [/tex]
[tex]t= \frac{285ln( \frac{1}{3}) }{ln( \frac{1}{2}) } [/tex]
[tex]t=451.71[/tex]

We can conclude that the correct answer is: a. about 452 days.