The antibiotic clarithromycin is eliminated from the body according to the formula
![A(t) = 500e^(-0.1386t)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/emmddu811tbl4ziny8xmillwu88birk9tf.png)
where A is the amount remaining in the body (in milligrams) t hours after the drug reaches peak concentration
We need to find the time 't' when amount of drug is reduced to 100
Plug in 100 for A(t) and solve for t
![100 = 500e^(-0.1386t)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/2w9g1eqhwhf8nwp8flh6v2j60etlprddsq.png)
Divide both sides by 500
![(1)/(5) = e^(-0.1386t)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7icf9is46rafsal7d0xey7uicqufvjlqtm.png)
Take ln on both sides
![ln((1)/(5)) = ln(e^(-0.1386t))](https://img.qammunity.org/2019/formulas/mathematics/middle-school/n4qw0nnda8no2c03vep7qro7ahyudd7y0t.png)
![ln((1)/(5)) = -0.1386tln(e)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qy9katv9khipq0pxndgdizo5120yt0cgyb.png)
![ln((1)/(5)) = -0.1386t](https://img.qammunity.org/2019/formulas/mathematics/middle-school/9ehxmfjlvnv8bexa4tf04h5nfevx1cfxh1.png)
divide both sides by -0.1386
t = 11.6121
11.61 hours will pass before the amount of drug in the body is reduced to 100.