Text A radioactive substance has initial mass 808080 g and halves every 555 years. (a) write the law y=b⋅axy=b\cdot a^{x}y=b⋅ax of mass versus time xxx (years); (b) compute the mass after 121212 years; (c) after how many years does the mass fall to 555 g? (use log102≈0,301\log_{10}2\approx0{,}301log102≈0,301) Solution (a) b=80b=80b=80 and, from halving in 555 years, a=(12)1/5=2−1/5≈0,8706a=\left(\tfrac12\right)^{1/5}=2^{-1/5}\approx0{,}8706a=(21)1/5=2−1/5≈0,8706: y=80⋅(12)x/5\;y=80\cdot\left(\tfrac12\right)^{x/5}y=80⋅(21)x/5. (b) y(12)=8022,4≈15,15y(12)=\dfrac{80}{2^{2,4}}\approx 15{,}15y(12)=22,480≈15,15 g. (c) 5=80⋅(12)x/5⇒(12)x/5=116=(12)4⇒x=205=80\cdot\left(\tfrac12\right)^{x/5}\Rightarrow \left(\tfrac12\right)^{x/5}=\tfrac{1}{16}=\left(\tfrac12\right)^4\Rightarrow x=205=80⋅(21)x/5⇒(21)x/5=161=(21)4⇒x=20 years. y=80⋅(12)x/5; y(12)≈15,15 g; x=20 years\boxed{y=80\cdot\left(\tfrac12\right)^{x/5};\ \ y(12)\approx15{,}15\ \text{g};\ \ x=20\ \text{years}}y=80⋅(21)x/5; y(12)≈15,15 g; x=20 years