Question

you take a 325 milligram dosage of ibuprofen. during each subsequent hour, the amount of medication in your bloodstream decreases by about 29% each hour.

a. write an exponential decay model giving the amount ( y ) (in milligrams) of ibuprofen in your bloodstream ( t ) hours after the initial dose. find and interpret the ( y )-value when ( t = 1.5 ). round your answer to the nearest whole number.

the exponential decay model is ( y = square )

when ( t = 1.5 ), the ( y )-value is about ( square ).

so, there are about (square) milligrams of ibuprofen in your bloodstream after (square)

b. estimate how long it takes for you to have 100 milligrams of ibuprofen in your bloodstream. round your answer to the nearest tenth.

it takes about (square) hours.

Explanation:

Step1: Recall exponential decay formula

The general form of an exponential decay model is \( y = a(1 - r)^t \), where \( a \) is the initial amount, \( r \) is the rate of decay (as a decimal), and \( t \) is time. Here, \( a = 325 \) mg, \( r = 0.29 \) (since 29% = 0.29). So the model is \( y = 325(1 - 0.29)^t = 325(0.71)^t \).

Step2: Calculate y when t = 1.5

Substitute \( t = 1.5 \) into the model: \( y = 325(0.71)^{1.5} \). First, calculate \( 0.71^{1.5} \). \( 0.71^{1.5}=\sqrt{0.71^3}=\sqrt{0.71\times0.71\times0.71}=\sqrt{0.71\times0.5041}=\sqrt{0.357911}\approx0.598 \). Then \( y = 325\times0.598\approx325\times0.6 = 195 \), but more accurately, \( 325\times0.598 = 325\times(0.6 - 0.002)=195 - 0.65 = 194.35\approx194 \). Wait, let's compute \( 0.71^{1.5} \) using a calculator approach: \( \ln(0.71)\approx - 0.3337 \), multiply by 1.5: \( - 0.50055 \), exponentiate: \( e^{-0.50055}\approx0.606 \). Then \( 325\times0.606 = 325\times0.6 + 325\times0.006 = 195 + 1.95 = 196.95\approx197 \)? Wait, maybe better to use \( 0.71^{1.5}=0.71\times\sqrt{0.71}\approx0.71\times0.8426\approx0.600 \). So \( 325\times0.600 = 195 \). Wait, let's do it properly: \( 0.71^{1.5}=e^{1.5\ln(0.71)}\approx e^{1.5\times(-0.3337)}=e^{-0.50055}\approx0.606 \). Then \( 325\times0.606 = 325\times0.6 + 325\times0.006 = 195 + 1.95 = 196.95\approx197 \). Wait, maybe my initial square root was wrong. Let's compute \( 0.71^3 = 0.71\times0.71 = 0.5041; 0.5041\times0.71 = 0.357911 \). Then square root of 0.357911: \( \sqrt{0.357911}\approx0.598 \). So \( 325\times0.598 = 325\times0.5 + 325\times0.098 = 162.5 + 31.85 = 194.35\approx194 \). Hmm, maybe the calculator is better. Let's use a calculator: \( 0.71^{1.5} \approx 0.71^{1 + 0.5}=0.71\times\sqrt{0.71}\approx0.71\times0.8426\approx0.600 \). So \( 325\times0.600 = 195 \). So the model is \( y = 325(0.71)^t \), and at \( t = 1.5 \), \( y\approx195 \) (rounded to nearest whole number).

Step3: Solve for t when y = 100

We have \( 100 = 325(0.71)^t \). Divide both sides by 325: \( \frac{100}{325}=(0.71)^t \). Simplify \( \frac{100}{325}=\frac{4}{13}\approx0.3077 \). Take natural logarithm of both sides: \( \ln(0.3077)=t\ln(0.71) \). \( \ln(0.3077)\approx - 1.179 \), \( \ln(0.71)\approx - 0.3337 \). So \( t=\frac{-1.179}{-0.3337}\approx3.53 \approx 3.5 \) hours.

Answer:

(part a model):
\( y = 325(0.71)^t \)