Question

homework assignment 6.7 exponential and logarithmic models
due friday by 11:59pm points 10 submitting an external tool
homework assignment 6.7 exponential and logarithmic models
score: 3/10 answered: 3/10
question 4
at the beginning of an experiment, a scientist has 348 grams of radioactive goo. after 150 minutes, her sample has decayed to 43.5 grams.
what is the half - life of the goo in minutes?
find a formula for ( g(t) ), the amount of goo remaining at time ( t ).
( g(t)=)
how many grams of goo will remain after 80 minutes?
you may enter the exact value or round to 2 decimal places.
question help: video read written example

Explanation:

Step1: Use the exponential decay formula

The general formula for exponential decay is \(G(t)=G_0e^{-kt}\), where \(G_0 = 348\) (initial amount), \(t = 150\) (time), and \(G(150)=43.5\). Substitute these values into the formula: \(43.5 = 348e^{-150k}\).

Step2: Solve for \(k\)

Divide both sides by \(348\): \(\frac{43.5}{348}=e^{-150k}\). Simplify \(\frac{43.5}{348}=\frac{1}{8}\), so \(\frac{1}{8}=e^{-150k}\). Take the natural logarithm of both sides: \(\ln(\frac{1}{8})=- 150k\). Since \(\ln(\frac{1}{8})=-3\ln(2)\), then \(-3\ln(2)=-150k\). Solve for \(k\): \(k=\frac{\ln(2)}{50}\).

Step3: Find the half - life \(T\)

For half - life, \(G(T)=\frac{G_0}{2}\). Using \(G(t)=G_0e^{-kt}\), substitute \(G(T)=\frac{G_0}{2}\): \(\frac{G_0}{2}=G_0e^{-kT}\). Divide both sides by \(G_0\): \(\frac{1}{2}=e^{-kT}\). Take the natural logarithm: \(\ln(\frac{1}{2})=-kT\). Since \(k = \frac{\ln(2)}{50}\), then \(-\ln(2)=-\frac{\ln(2)}{50}T\). Solve for \(T\): \(T = 50\) minutes.

Step4: Write the formula for \(G(t)\)

Substitute \(G_0 = 348\) and \(k=\frac{\ln(2)}{50}\) into \(G(t)=G_0e^{-kt}\), we get \(G(t)=348e^{-\frac{\ln(2)}{50}t}=348\times2^{-\frac{t}{50}}\).

Step5: Find the amount after \(t = 80\) minutes

Substitute \(t = 80\) into \(G(t)=348\times2^{-\frac{t}{50}}\): \(G(80)=348\times2^{-\frac{80}{50}}=348\times2^{-\frac{8}{5}}\). Calculate \(2^{-\frac{8}{5}}=\frac{1}{2^{\frac{8}{5}}}=\frac{1}{\sqrt[5]{256}}\approx0.378\). Then \(G(80)=348\times0.378\approx131.54\) grams.

Answer:

Half - life: \(50\) minutes.
Formula: \(G(t)=348\times2^{-\frac{t}{50}}\).
Amount after \(80\) minutes: \(131.54\) grams.