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
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$\begingroup$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.
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