The weight (W kg) of a decaying radio active substance after n years is given by W= Wo(1/2)^n/100, where Wo kg is the initial weight of the substance. 1. Atleast how many years will it take for the radioactive substance to lose to 10% of its initial weight?

[tex]\begin{gathered} \frac{10}{100}=\frac{100}{100}\mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \\ \implies0.1=\mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \end{gathered}[/tex]

We then solve the equation for the value of n.

Take the logarithm of both sides.

[tex]\begin{gathered} \log (0.1)=\log \mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \\ \implies\log (0.1)=\frac{n}{100}\log (\frac{1}{2})^{} \end{gathered}[/tex]

Then divide both sides by log(1/2):

[tex]\begin{gathered} \frac{\log (0.1)}{\log (\frac{1}{2})}=\frac{\frac{n}{100}\log(\frac{1}{2})^{}}{\log(\frac{1}{2})} \\ \implies\frac{n}{100}=\frac{\log (0.1)}{\log (\frac{1}{2})} \end{gathered}[/tex]

Finally, multiply both sides by 100:

[tex]\begin{gathered} 100\times\frac{n}{100}=100\times\frac{\log (0.1)}{\log (\frac{1}{2})} \\ n=332.19\text{ years} \end{gathered}[/tex]

It will take at least 332.19 years for the radioactive substance to lose to 10% of its initial weight.