Exponential Growth and Decay

1. Table A: Check ratios of consecutive $y$-values:

$\frac{6}{3} = 2, \quad \frac{12}{6} = 2, \quad \frac{24}{12} = 2$

The ratio is constant ($\times 2$ each time). Table A is exponential: $y = 3 \cdot 2^x$.

Table B: Check differences of consecutive $y$-values:

$8 - 5 = 3, \quad 11 - 8 = 3, \quad 14 - 11 = 3$

The difference is constant ($+3$ each time). Table B is linear: $y = 3x + 5$.

2. (a) $f(x) = 500 \cdot (1.06)^x$

(b) Halving each year means the growth factor is $\frac{1}{2}$:

$f(x) = 200 \cdot \left(\frac{1}{2}\right)^x$

3. (a) Growth rate $4\% = 0.04$, so the growth factor is $1 + 0.04 = 1.04$:

$P(t) = 8000 \cdot (1.04)^t$

(b) 2020 is $t = 10$ years after 2010:

$P(10) = 8000 \cdot (1.04)^{10} \approx 8000 \cdot 1.4802 \approx 11{,}842$

The projected population is approximately 11,842 people.

(c) Find when $P(t) > 12{,}000$:

• $P(10) \approx 11{,}842 < 12{,}000$
• $P(11) = 8000 \cdot (1.04)^{11} \approx 8000 \cdot 1.5395 \approx 12{,}316 > 12{,}000$

The population first exceeds $12{,}000$ at $t = 11$, i.e., in 2021.

4. (a) Depreciation rate $15\% = 0.15$, so the decay factor is $1 - 0.15 = 0.85$:

$V(t) = 24000 \cdot (0.85)^t$

(b) After $3$ years:

$V(3) = 24000 \cdot (0.85)^3 = 24000 \cdot 0.614125 \approx \$14{,}739$

(c) After $5$ years:

$V(5) = 24000 \cdot (0.85)^5 = 24000 \cdot 0.4437 \approx \$10{,}649$

5. (a) At $x = 10$:

$f(10) = 50(10) = 500$ $g(10) = 2^{10} = 1024$

Function B ($g$) is greater at $x = 10$.

(b) At $x = 20$:

$f(20) = 50(20) = 1000$ $g(20) = 2^{20} = 1{,}048{,}576$

Function B ($g$) is vastly greater at $x = 20$.

(c) Even though a linear function may start larger, exponential growth eventually and permanently overtakes linear growth. Exponential functions grow by a constant factor each step, while linear functions grow by a constant amount — the factor-based growth compounds and accelerates without bound.

6. (a) Every $5$ years, mass is multiplied by $\frac{1}{2}$. In $t$ years, there are $\frac{t}{5}$ half-lives:

$M(t) = 640 \cdot \left(\frac{1}{2}\right)^{t/5}$

(b) After $15$ years: $\frac{15}{5} = 3$ half-lives:

$M(15) = 640 \cdot \left(\frac{1}{2}\right)^3 = 640 \cdot \frac{1}{8} = 80 \text{ grams}$

(c) After $25$ years: $\frac{25}{5} = 5$ half-lives:

$M(25) = 640 \cdot \left(\frac{1}{2}\right)^5 = 640 \cdot \frac{1}{32} = 20 \text{ grams}$