Functions - Exponential functions

The population of a city is modeled by the function $P(t) = 12000(1.035)^t$, where $t$ is the number of years after 2010. Find the population in 2025 and determine the annual percentage growth rate.

1. Identify the time interval: $t = 2025 - 2010 = 15$ years.
2. Substitute $t = 15$ into the function: $P(15) = 12000(1.035)^{15}$ $P(15) \approx 12000 \times 1.6753$ $P(15) \approx 20104$
3. Identify the growth rate from the base $b = 1.035$: $1 + \frac{r}{100} = 1.035$ $\frac{r}{100} = 0.035$ $r = 3.5\%$

To find the value at a specific time, we substitute the value of $t$. The base $1.035$ represents a $3.5\%$ increase per year because $1.035 = 1 + 0.035$.

An exponential function passes through the points $(0, 8)$ and $(3, 64)$. Find the equation in the form $y = a \cdot b^x$.

1. Use the point $(0, 8)$ to find $a$: $8 = a \cdot b^0$ Since $b^0 = 1$, $a = 8$.
2. Use the point $(3, 64)$ and the value of $a$ to find $b$: $64 = 8 \cdot b^3$ $b^3 = \frac{64}{8}$ $b^3 = 8$ $b = \sqrt[3]{8}$ $b = 2$
3. Write the final equation: $y = 8 \cdot 2^x$

The $y$-intercept $(0, a)$ directly gives the initial value $a$ when there is no vertical shift. We then solve for the base $b$ using the second coordinate.

A car is purchased for 25000 and depreciates at a rate of $12\%$ per year. Write an expression for the value $V$ of the car after $n$ years and find its value after 5 years.

1. Identify the parameters: $a = 25000$ and $r = 12$.
2. Calculate the decay factor $b$: $b = 1 - \frac{12}{100} = 0.88$
3. Formulate the equation: $V = 25000(0.88)^n$
4. Calculate for $n = 5$: $V = 25000(0.88)^5$ $V \approx 25000 \times 0.5277$ $V \approx 13193.29$

Depreciation is modeled as exponential decay. The base is $(1 - \text{rate})$. The value after 5 years is calculated by raising the base to the power of 5 and multiplying by the initial price.

1. The general form is $B(t) = a \cdot b^{\frac{t}{c}}$, where $c$ is the doubling period.
2. Here, $a = 500$, $b = 2$, and $c = 3$.
3. Equation: $B(t) = 500 \cdot 2^{\frac{t}{3}}$
4. For $t = 12$: $B(12) = 500 \cdot 2^{\frac{12}{3}}$ $B(12) = 500 \cdot 2^4$ $B(12) = 500 \cdot 16 = 8000$ After 12 hours, there are 8000 bacteria.

To model doubling time, we use the base $2$ and divide the time $t$ by the duration it takes to double. Substituting the known values into this power function gives the total count.

1. The base $b = 0.85$.
2. Decay rate $r$: $b = 1 - \frac{r}{100} \implies 0.85 = 1 - \frac{r}{100} \implies r = 15\%$.
3. Calculate mass for $t = 10$: $M(10) = 100(0.85)^{10}$ $M(10) \approx 100(0.19687)$ $M(10) \approx 19.69\text{ g}$ The daily decay rate is $15\%$ and the remaining mass is $19.69\text{ g}$.

The base $0.85$ represents the proportion remaining; subtracting this from 1 gives the proportion lost (decay rate). We use the power of 10 to find the substance remaining after 10 full decay periods.