We have that the general expression for a function with exponential growth or decay is:

where 'a' represents the initial value, 'b' represents the growth or decay and t represents the time.
In this case, we have that a = 100, since that is the initial population (100 mg of radioactive substance). Also, since we have that after 9 hours, 55 mg of the substance remains, we have the following equation:

now, given the exponential function, with a = 100, we have:

then, combining both expressions, we get:

solving for b, we have:
![\begin{gathered} 100b^9=55 \\ \Rightarrow b^9=(55)/(100) \\ \Rightarrow b=\sqrt[9]{(55)/(100)}=0.935 \\ b=.935 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mv3btuxejag78levh5tbsxtq4vkinxdo85.png)
then, the function is defined as follows:

finally, to find out how many milligrams will remain after 21 hours, we can make t = 21 and evaluate the function:

therefore, there will be 24.38mg after 21 hours