Honors Algebra 2 - Unit 6: Exponentials, Logarithms, and Geometric Series

Ice Decay Modeling:   - The amount of ice remaining in a glass is represented by the exponential decay model:
$y = 12(0.85)^x$
    where:
    - $y$ is the amount of ice remaining in ounces.
    - $x$ is the time in minutes.   - Interpretation of Constants:     - The number 12 represents the initial amount of ice remaining in the glass (in ounces).     - The number 0.85 represents the decay factor; it indicates that the amount of ice decreases by $15\%$ per minute.   - Calculation Example:
    To determine how many ounces of ice remain after 20 minutes ( $x = 20$ ):
$y = 12(0.85)^{20}$
$y \approx 0.465\,\text{ounces}$
    Result: There will be approximately $0.465\,\text{ounces}$ of ice remaining.

Function Analysis: $y = 2^{x+1} - 3$
Horizontal Asymptote: $y = -3$
Key Characteristics: - Domain: $(-\infty, \infty)$ - Range: $(-3, \infty)$
Coordinate Point Mapping: - When $x = 0$ , $y = 2^{0+1} - 3 = 2 - 3 = -1$ ; point: $(0, -1)$ - When $x = 1$ , $y = 2^{1+1} - 3 = 4 - 3 = 1$ ; point: $(1, 1)$ - When $x = 2$ , $y = 2^{2+1} - 3 = 8 - 3 = 5$ ; point: $(2, 5)$ - When $x = -1$ , $y = 2^{-1+1} - 3 = 1 - 3 = -2$ ; point: $(-1, -2)$

Form: $y = ab^x$
Given Points: $(-2, 1/2)$ and $(0, 2)$ .
Step-by-Step Procedure:   1. Use the point $(0, 2)$ to find $a$ :
$2 = a \times b^0$
$2 = a \times 1 \implies a = 2$   2. Use the point $(-2, 1/2)$ and $a = 2$ to find $b$ :
$1/2 = 2 \times b^{-2}$
$1/4 = b^{-2}$
$1/4 = 1/b^2 \implies b^2 = 4 \implies b = \pm 2$ (base must be positive, so $b = 2$ ).   3. Resulting Function: $F(x) = 2(2)^x$

Scenario: You were sneezed on during a math activity, resulting in the deposition of 250 cold viruses. The virus increases by half of its amount (growth of $50\%$ ) every half hour (30 minutes).
Equation of Best Fit:
$F(x) = 250(1 + 0.5)^{\frac{x}{30}}$
where $x$ represents the time in minutes.
Calculation after 7.5 Hours:
1. Convert hours to minutes: $7.5\,\text{hours} \times 60\,\text{min/hr} = 450\,\text{minutes}$ . 2. Insert into the formula:
$F(450) = 250(1.5)^{450/30}$
$F(450) = 250(1.5)^{15}$
$F(450) \approx 109,473\,\text{viruses}$ . - Result: There will be approximately 109,473 viruses in your body after 7 and a half hours.

Condensing Expressions: - Single Logarithm 1: $[\log_w(y) + \log_w(z)] - 3\log_w(x)$
$\rightarrow \log_w(yz) - \log_w(x^3) = \log_w\left(\frac{yz}{x^3}\right)$ - Single Logarithm 2: $7\log(x) + \log(x+1) - 5\log(x)$
$\rightarrow 2\log(x) + \log(x+1) = \log(x^2) + \log(x+1) = \log(x^2(x+1))$
Expanding Expressions: - Expression: $\log\left(\frac{\sqrt{2xy}}{3}\right)$
$\rightarrow \log(\sqrt{2xy}) - \log(3)$
$\rightarrow \frac{1}{2}[\log(2) + \log(x) + \log(y)] - \log(3)$
Simplifying Expressions: - Expression: $3\log_2(4) - \log_2(16)$
Since $\log_2(4) = 2$ and $\log_2(16) = 4$ :
$3(2) - 4 = 6 - 4 = 2$

Solving for x (Exact Form):   1. Logarithmic Equation with Base Change:       $2\log_8(x) - \log_8(4) = 4$
     Note that $\log_8(4) = \frac{2}{3}$ because $8^{2/3} = 4$ :
$2\log_8(x) - \frac{2}{3} = 4$
$2\log_8(x) = \frac{14}{3} \implies \log_8(x) = \frac{7}{3}$
$x = 8^{7/3} = (2^3)^{7/3} = 2^7 = 128$   2. Linear Logarithmic Comparison:       $\log(7x+1) = \log(x-2) + 1$
$\log(7x+1) - \log(x-2) = 1$
$\log\left(\frac{7x+1}{x-2}\right) = 1$
$\frac{7x+1}{x-2} = 10^1$
$7x+1 = 10(x-2) = 10x - 20$
$21 = 3x \implies x = 7$   3. Natural Logarithm Equation:       $\ln(x^2-9) - \ln(x+3) = 5$
$\ln\left(\frac{x^2-9}{x+3}\right) = 5$
$\ln\left(\frac{(x-3)(x+3)}{x+3}\right) = 5$
$\ln(x-3) = 5$
$x-3 = e^5 \implies x = e^5 + 3$   4. Exponential Equation with e:       $3e^{3x} + 7 = 52$
$3e^{3x} = 45 \implies e^{3x} = 15$
$3x = \ln(15) \implies x = \frac{\ln(15)}{3}$

Population Growth (RV School): - Current buildings capacity: $2,500\,\text{students}$ . - Current population: $2,150\,\text{students}$ increasing by $2\%\,\text{each year}$ . - Model: $2500 = 2150(1.02)^t$ - Solution:
$\frac{2500}{2150} = (1.02)^t$
$\log\left(\frac{50}{43}\right) = t \log(1.02)$
$t = \frac{\log(1.16279)}{\log(1.02)} \approx 7.6\,\text{years}$ . - Result: A new school will be needed in about 8 years.
Compound Interest (Joe's Bank Account): - Account Stats: Current balance $A = \$4,200$ . Deposited 5 years ago ( $t=5$ ) at $4.75\%\,\text{interest}$ compounded monthly ( $n=12$ ). - Initial Deposit calculation: $4200 = P\left(1 + \frac{0.0475}{12}\right)^{12 \times 5}$ $4200 = P(1.0039583)^{60}$ $P = \frac{4200}{1.26627} \approx \$3,313.66$ - Doubling Time calculation: Find how long it takes for the initial deposit to double ( $2P = P(1 + r/n)^{nt}$ ): $2 = \left(1 + \frac{0.0475}{12}\right)^{12t}$ $\log(2) = 12t \log\left(1 + \frac{0.0475}{12}\right)$ $t = \frac{\log(2)}{12 \log(1.003958)} \approx 14.621\,\text{years}$ . - Result: It will take about 15 years from the initial deposit for the money to double.

Sequence Problem: $11, 1.1, 0.11, \dots$ - nth Term: $a_n = 11(0.1)^{n-1}$ - Sum of first 20 terms ( $S_{20}$ ):
Using $S_n = \frac{a_1(1-r^n)}{1-r}$ : $S_{20} = \frac{11(1 - 0.1^{20})}{1 - 0.1} = \frac{11(1)}{0.9} \approx 12.222$
Determining number of terms (n): - Given: $a = 4, r = 2, S_n = 124$ - Solution:
$124 = \frac{4(1-2^n)}{1-2}$
$124 = \frac{4(1-2^n)}{-1} = -4(1-2^n) = 4(2^n-1)$ $31 = 2^n - 1 \implies 32 = 2^n$ $2^5 = 2^n \implies n = 5$

Series 1: $\sum_{n=1}^{\infty} \frac{-5}{4^{n-1}}$ - Variables: $a_1 = -5, r = 1/4$ - State: Convergent ( $|r| < 1$ ). - Sum: $S_{\infty} = \frac{-5}{1 - 1/4} = \frac{-5}{3/4} = \frac{-20}{3}$ .
Series 2: $\sum_{n=1}^{\infty} 2^n(5)^{n-1}$ - Variables: $a_1 = 2, r = 10$ (because $\sum 2 \times 2^{n-1} \times 5^{n-1} = \sum 2(10)^{n-1}$ ). - State: Divergent ( $|r| > 1$ ).
Series 3: $\sum_{n=0}^{\infty} \frac{4^{n+2}}{2^{3-4n}}$ - Simplification: According to the student work, when $n=0$ , current term is $2$ ; when $n=1$ , current term is $128$ (wait, student notes contradict convergence, but indicate $a_1 = 128, r = 1/4$ ). - State: Convergent (Student claim). - Sum: $S_{\infty} = \frac{128}{1 - 1/4} = \frac{128}{3/4} = \frac{512}{3}$ (Student calculation).