Solution:

Step 1: Identify the initial value and growth pattern.

• Initial population: $A_0 = 200$
• The population doubles every 3 hours

Step 2: Determine the base for doubling every 3 hours.

Since it doubles every 3 hours, after 3 hours we have $2 \times 200 = 400$.

We need to find $b$ such that $200b^3 = 400$.

$b^3 = 2$ $b = 2^{1/3} = \sqrt[3]{2}$

Step 3: Write the function.

$A(t) = 200 \cdot 2^{t/3}$

OR equivalently: $A(t) = 200 \cdot (\sqrt[3]{2})^t$

Step 4: Verify.

• At $t = 0$: $A(0) = 200 \cdot 2^{0/3} = 200$ ✓
• At $t = 3$: $A(3) = 200 \cdot 2^{3/3} = 200 \cdot 2 = 400$ ✓
• At $t = 6$: $A(6) = 200 \cdot 2^{6/3} = 200 \cdot 4 = 800$ ✓

Answer: $A(t) = 200 \cdot 2^{t/3}$ bacteria

Solution:

Part (a): For exponential growth with doubling, we use: $P(t) = P_0 \cdot 2^{t/T}$

where:

• $P_0 = 500$ (initial population)
• $T = 3$ (doubling time in hours)

$P(t) = 500 \cdot 2^{t/3}$

Part (b): After 12 hours:

$P(12) = 500 \cdot 2^{12/3} = 500 \cdot 2^4 = 500 \cdot 16 = 8000$ bacteria

Part (c): Set $P(t) = 16000$ and solve:

$16000 = 500 \cdot 2^{t/3}$

$32 = 2^{t/3}$

$2^5 = 2^{t/3}$

$5 = \frac{t}{3}$

$t = 15$ hours

Solution:

Step 1: Identify the initial value and decay rate.

• Initial value: $A_0 = 25000$
• Decay rate: $r = 0.15$ (15%)

Step 2: Write the exponential decay model.

Since the car loses 15% each year, it retains 85% of its value: $A(t) = 25000(1 - 0.15)^t = 25000(0.85)^t$

Step 3: Calculate the value after 5 years.

$A(5) = 25000(0.85)^5$ $A(5) = 25000(0.4437...)$ $A(5) \approx 11,092.63$

Answer: The car is worth approximately $11,093 after 5 years.

Solution:

Part (a): The formula $A(t) = A_0 e^{kt}$ has decay rate $|k|$.

Here $k = -0.0315$, so the decay rate is $0.0315 = 3.15\%$ per year.

Part (b): Half-life is when $A(t) = \frac{A_0}{2}$:

$\frac{A_0}{2} = A_0 e^{-0.0315t}$

$\frac{1}{2} = e^{-0.0315t}$

$\ln(0.5) = -0.0315t$

$t = \frac{\ln(0.5)}{-0.0315} = \frac{-0.693}{-0.0315} \approx 22.0$ years

Part (c): With $A_0 = 200$ grams and $t = 50$ years:

$A(50) = 200 \cdot e^{-0.0315(50)}$

$A(50) = 200 \cdot e^{-1.575}$

$A(50) = 200 \cdot 0.2067$

$A(50) \approx 41.3$ grams

Solution:

Part a) $f(x) = 5(1.2)^x$

Base: $b = 1.2 > 1$

Since the base is greater than 1, this represents exponential growth.

Part b) $g(x) = 3(0.8)^x$

Base: $b = 0.8 < 1$ (but $0.8 > 0$)

Since the base is between 0 and 1, this represents exponential decay.

Part c) $h(x) = 2e^{-0.5x}$

Rewrite: $h(x) = 2(e^{-0.5})^x$

The base is $e^{-0.5} \approx 0.606 < 1$

Since the base is between 0 and 1, this represents exponential decay.

Alternative approach for part c): The negative exponent indicates decay.

Answers:

• a) Growth
• b) Decay
• c) Decay