A 737.8 ng sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 47 days the amount of radioactive substance had decreased to 92.23 ng. How many half-lives of the unknown radioactive substance have occurred?

where N(t) is the amount of the substance that remains after t time, [tex]N_0[/tex] is the initial amount, t is the time that elapses and [tex]t_H[/tex] is the amount of time it takes the substance to half its size.

Understanding the Exponent

The [tex]\dfrac{t}{t_H}[/tex] represents how many times the substance's quantity is halved or how many half-lives it experiences

Both t and [tex]t_H[/tex] must have the same metric of time, either than that, it can be in terms of,

seconds
minutes
hours
days
etc.

For example,

if element A has a half life of 6 years and 12 years elapses then, element A halves itself twice or experiences two half-lives.

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Solving the Problem

The problem asks for the number of half-lives the given sample had in a course of 47 days, meaning that we must solve for the exponent in the equation the word problem makes.

[tex]92.23=737.8\left(\dfrac{1}{2} \right)^T[/tex],

let T be the number of half-lives that this substance experiences, writing out the exponent is unnecessary since the problem doesn't ask for the value of [tex]t_H[/tex] nor each element in the exponent's fraction.

Now we rearrange and isolate the T variable.

[tex]\dfrac{92.23}{737.8} =\left(\dfrac{1}{2}\right)^T[/tex]

[tex]0.125=\dfrac{1}{8} =\left(\dfrac{1}{2}\right)^T[/tex]

Knowing that,

[tex]log_{\dfrac{1}{2} } \left(\dfrac{1}{8} \right)=3[/tex]

or that,

[tex]\left(\dfrac{1}{2} \right)^3=\dfrac{1^3}{2^3} =\dfrac{1}{8}\\[/tex]

T = 3.

So, after 47 days the substances experiences 3 half-lives.