Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an 18.0 g sample after 60 hours?

Using

A = 18e^(kt)... where A is the amount after t hours,  we have

9 = 18e^(k*15)    divide both sides by 18

1/2 = e^(15*k)    take the ln of both sides

ln(1/2) = ln e^(15* k)   and by a log property, we can write

ln(1/2) = (15k) lne    and lne = 1    so we can drop that part

ln(1/2) = 15k        divide both sides by 15

k = ln(1/2) / 15 = -.0462

So after 60 hours, we have

A = 18e^(60* -0462)  = about 1.126 g

Here's the graph of the situation.....where x = t   ....the relevant values are where x ≥ 0

https://www.desmos.com/calculator/wvzugktxmp