Respuesta :
We have the equation:
[tex]W=W_0(\frac{1}{2})^{\frac{n}{100}}[/tex]And we want to find the value n, correspondign to the number of years necessary in order to the substance to decay in half.
Let's say that we have 1 Kg of the substance, this is Wo, the initial weight. Since we want to find the the decay of half the substance we use W = 1/2
And write:
[tex]\frac{1}{2}=1\cdot(\frac{1}{2})^{\frac{n}{100}}[/tex]Now we can use a property of logarithms:
[tex]\ln (a^b)=b\ln (a)[/tex]Thus applying natural log on both sides:
[tex]\ln (\frac{1}{2})=\ln (\frac{1}{2}^{\frac{n}{100}})[/tex]By the property:
[tex]\ln (\frac{1}{2})=\frac{n}{100}\ln (\frac{1}{2}^{})[/tex]We can divide on both sides by ln(1/2):
[tex]\begin{gathered} 1=\frac{n}{100} \\ n=100 \end{gathered}[/tex]The number of years for the radioactive substance to decay to half its initial weigh are 100 years.
The step to get rid of the ln(1/2) is:
[tex]\begin{gathered} \ln (\frac{1}{2})=\frac{n}{100}\ln (\frac{1}{2}^{}) \\ \frac{\ln (\frac{1}{2})}{\ln (\frac{1}{2})}=\frac{n}{100}\frac{\ln (\frac{1}{2}^{})}{\ln (\frac{1}{2})} \\ 1=\frac{n}{100}\cdot1 \\ 1=\frac{n}{100} \end{gathered}[/tex]
