If there are 24 hours in a day, how can you write the number of days that have passed in terms of the number of hours $h$ that have passed?

Once you have an expression for $d$ in terms of $h$, substitute it into the exponent $d/18$. Then simplify the resulting fraction in the exponent carefully before comparing to the answer choices.

In Desmos, type M0 = 1 to fix an initial mass, then enter the function M(d) = M0*(1/2)^(d/18) so you have the original decay model in terms of days.

Type four functions: A1(h) = M0*(1/2)^(h/18), A2(h) = M0*(1/2)^(h/(18*24)), A3(h) = M0*(1/2)^(18*h), and A4(h) = M0*(2)^(h/(18*24)) corresponding to choices A–D.

Pick a convenient time like 24 hours (which is 1 day). In Desmos, evaluate M(1) and also A1(24), A2(24), A3(24), and A4(24) (you can type each directly or use a table). The correct choice is the one whose value at 24 hours matches the value of the original model at 1 day.

The function $M(d)=M_0\bigl(\tfrac{1}{2}\bigr)^{d/18}$ gives the mass $d$ days after the initial measurement.

• The exponent $d/18$ means that every 18 days, the mass is multiplied by $\tfrac{1}{2}$ (it has a half-life of 18 days).
• Right now, the input (the variable) is in days, not hours.

We want $A(h)$, the mass $h$ hours after the initial measurement.

• There are 24 hours in 1 day.
• If $h$ is the number of hours that have passed, then the number of days that have passed is

So $d$ (days) and $h$ (hours) are connected by $d = h/24$.

Since $A(h)$ should give the same mass as $M(d)$, just measured at a time of $h$ hours, we replace $d$ with $h/24$ in the original formula:

This is now a function of $h$ (hours), but the exponent can be simplified further.

Simplify the exponent $(h/24)/18$:

So the hourly model is

which matches answer choice B.