Exponential growth and decay: 3 worked examples

The population growth can be modeled using the formula for exponential growth: $P = P_0(1 + r)^t$, where $P_0$ is the initial population, $r$ is the growth rate as a decimal, and $t$ is the number of years.
Given:
Initial population $P_0 = 8,000$
Growth rate $r = 5\% = 0.05$
Time $t = 6$ years

Substituting the values into the formula:
$P = 8,000(1 + 0.05)^6$
$P = 8,000(1.05)^6$

First, calculate $(1.05)^6$:
$(1.05)^1 = 1.05$
$(1.05)^2 = 1.1025$
$(1.05)^3 = 1.157625$
$(1.05)^6 \approx (1.157625)^2 \approx 1.3400956$

Now, multiply by the initial population:
$P = 8,000 \times 1.3400956$
$P = 10,720.7648$

Rounding to the nearest person, we get $10,721$.

The value of the car follows an exponential decay model given by the formula $V = P(1 - r)^{t}$, where $P$ is the initial principal value, $r$ is the rate of depreciation, and $t$ is the time in years.

Given:
$P = 24,000$
$r = 0.15$
$t = 4$

Substituting the values into the formula:
$V = 24,000(1 - 0.15)^{4}$
$V = 24,000(0.85)^{4}$

Calculating $(0.85)^{4}$:
$(0.85)^{2} = 0.7225$
$(0.7225)^{2} = 0.52200625$

Now, multiply by the initial value:
$V = 24,000 \times 0.52200625$
$V = 12,528.15$

Rounding to the nearest dollar, we get $12,528.

The formula for radioactive decay based on half-life is $A = A_{0} \cdot (\frac{1}{2})^{\frac{t}{h}}$, where $A_{0}$ is the initial amount, $t$ is the elapsed time, and $h$ is the half-life period.
Given:
$A_{0} = 500$ grams
$t = 24$ hours
$h = 8$ hours

First, determine the number of half-life cycles that have passed:
$n = \frac{t}{h} = \frac{24}{8} = 3$ cycles.

Now, calculate the remaining amount:
$A = 500 \cdot (\frac{1}{2})^{3}$
$A = 500 \cdot \frac{1}{8}$
$A = 62.5$

After 24 hours, 62.5 grams of the substance remain.