Let the initial mass be represented by
. Therefore,
.
Let the final mass (or present mass be represented by
. Therefore,
.
Now, we are given that the rate of decay is 8% or 0.008 per day. Therefore, the amount of substance remaining after each day will be
. Let us represent this by
.
Please note that the equation guiding such decays is always given by:
, where t is the time.
Thus, the above equation will give in our case:
![115=240(0.92)^t](https://img.qammunity.org/2019/formulas/mathematics/high-school/57t0eihvvhrfy9nub456ux4p64sci1w2xc.png)
![(115)/(240)\approx0.4792=(0.92)^t](https://img.qammunity.org/2019/formulas/mathematics/high-school/r8v3ocpd9spswco5o6w5c9kwfpgi86scgq.png)
Taking natural logarithm on both sides we get:
![ln(0.4792)=t* ln(0.92)](https://img.qammunity.org/2019/formulas/mathematics/high-school/c40ifflr7ydr7kx6qogkutmr4mlws1yhxe.png)
Therefore,
![t=(ln(0.4792))/(ln(0.92))](https://img.qammunity.org/2019/formulas/mathematics/high-school/haa2p5ane3jbz3dc5oi2302th2qpvz7826.png)
days<9 days
Thus, in 9 days there will be further decay and the sample left will definitely be less than 115 milligrams.
The only option which matches this reality is Option C. Therefore, Option C is the correct answer.