The radioactive polonium decays to half of its original amount every 139 days (i.e. its half-life is 139 days). If your sample will not be useful to you after 78% of the radioactive nuclei present on the day the sample arrives had disintegrated, for about how many days after the sample arrives will you be able to use your polonium sample?
I tried using formula f(X) = a*b^h, then I am getting negative value for h , which I can't solve further using log or ln
You set up an exponential function based on days and percentage left. You know that the point (0, 1) will be on the graph because at 0 days it is 100% there. Then after 139 days, the amount will decrease to 1/2 of its original amount so you have the point (139, 1/2). And (278, 1/4) and so on. Then you know that in the formula $f(x) = b*a^{x}$, b=1 because were given the intercept in the point (0, 1). Then to get a, you say that $1/2 = a^{139}$ therefore $a = (1/2)^{1/139}$ so $f(x) = (1/2)^{x/139}$. Then if you want to know when 78% of it is gone (i.e. you have 22% left) you would say you want to solve for x when f(x) is 22/100. That is $11/50 = (1/2)^{x/139}$ therefore $\log_{1/2}(11/50) = x/139$ and finally $x = 139*\log_{1/2}(11/50)$ and after x days it goes bad.
