Question

a scientist is studying how bacteria thrive under different conditions of temperature and pressure. the scientist develops bacteria population models for the different scenarios.
condition 1: $p(t)=500(0.3)^t$
condition 2: $p(t)=500(1.65)^{\frac{t}{4}}$
condition 3: $p(t)=500(0.95)^{2t}$
condition 4: $p(t)=500(1.2)^{0.5t}$
drag and drop each population model under the column that describes the change in bacteria population over time.
population growth | population decay
$p(t)=500(0.3)^t$ $p(t)=500(1.65)^{\frac{t}{4}}$ $p(t)=500(0.95)^{2t}$ $p(t)=500(1.2)^{0.5t}$

Explanation:

Step1: Recall exponential - growth and decay rules

For an exponential function $P(t)=a\cdot b^{kt}$, if $b > 1$, it represents growth; if $0 < b<1$, it represents decay.

Step2: Analyze $P(t)=500(0.3)^{t}$

Here $b = 0.3$ and $0<0.3<1$, so it represents population decay.

Step3: Analyze $P(t)=500(1.65)^{\frac{t}{4}}$

Here $b = 1.65$ and $1.65>1$, so it represents population growth.

Step4: Analyze $P(t)=500(0.95)^{2t}$

Let $u = 2t$, then the function is $P(u)=500(0.95)^{u}$. Since $0<0.95<1$, it represents population decay.

Step5: Analyze $P(t)=500(1.2)^{0.5t}$

Let $v=0.5t$, then the function is $P(v)=500(1.2)^{v}$. Since $1.2 > 1$, it represents population growth.

Answer:

Population Growth	Population Decay