Detailed Study Notes: Exponential and Logarithmic Functions

Solving Exponential Functions

• General Concepts and Methodology:     * To solve exponential equations where the bases can be made the same, use the One-to-One Property: if $b^x = b^y$, then $x = y$.     * When bases cannot easily be match, use logarithms or natural logarithms ($\ln$) to bring the exponent down using the Power Property.     * Isolate the exponential term before applying logarithmic operations.

• Specific Problems and Formulations:     1. $2^{x+2} = 25$         * This requires converting to logarithmic form or taking the log of both sides: $\log_{2}(25) = x + 2$ or $(x+2)\text{ln}(2) = \text{ln}(25)$.     2. $4^{5x} = 4^{2x+6}$         * Applying the One-to-One Property: $5x = 2x + 6$.         * Solving for $x$: $3x = 6 \implies x = 2$.     3. $3(2)^{x+1} = 48$         * Step 1: Divide by $3$ to isolate the base: $2^{x+1} = 16$.         * Step 2: Recognize $16$ as a power of $2$: $2^{x+1} = 2^4$.         * Step 3: Equate exponents: $x + 1 = 4 \implies x = 3$.     4. $e^{x+2} = 8$         * Take the natural log ($\ln$) of both sides: $\text{ln}(e^{x+2}) = \text{ln}(8)$.         * Simplify: $x + 2 = \text{ln}(8)$.         * Solve: $x = \text{ln}(8) - 2$.     5. $2(5)^{x-3} + 4 = 54$         * Step 1: Subtract $4$ from both sides: $2(5)^{x-3} = 50$.         * Step 2: Divide by $2$: $5^{x-3} = 25$.         * Step 3: Express $25$ as $5^2$: $5^{x-3} = 5^2$.         * Step 4: Equate exponents: $x - 3 = 2 \implies x = 5$.     6. $3(2)^{x-3} - 4 = 53$         * Step 1: Add $4$ to both sides: $3(2)^{x-3} = 57$.         * Step 2: Divide by $3$: $2^{x-3} = 19$.         * Step 3: Use logs: $(x-3)\text{ln}(2) = \text{ln}(19)$.

Exponential Application Problems

• Compound Interest (Periodic):     * Formula: $A = P(1 + \frac{r}{n})^{nt}$     * Scenario: A $\$5,000$ investment at a rate of $5\%$ compounded monthly for eight years.     * Variables:         * Principal ($P$) = $\$5,000$         * Annual Interest Rate ($r$) = $0.05$         * Compounding periods per year ($n$) = $12$         * Time ($t$) = $8$     * Calculation: $A = 5000(1 + \frac{0.05}{12})^{12 \times 8}$

• Depreciation (Exponential Decay):     * Formula: $V = P(1 - r)^t$     * Scenario: A car bought for $\$20,000$ depreciating at $13\%$ per year. Goal: Determine when value hits $\$12,000$.     * Variables:         * Principal ($P$) = $\$20,000$         * Rate of Depreciation ($r$) = $0.13$         * Ending Value ($V$) = $\$12,000$     * Equation: $12,000 = 20,000(1 - 0.13)^t$     * Solving for $t$: $0.6 = (0.87)^t \implies t = \frac{\text{ln}(0.6)}{\text{ln}(0.87)}$

• Population Growth (Doubling Time):     * Formula: $N = N_0(2)^{\frac{t}{d}}$     * Scenario: A virus colony initially has $50,000$ cell, doubling every $8$ hours. Goal: Size in two days.     * Variables:         * Initial Population ($N_0$) = $50,000$         * Time ($t$) = $2$ days = $48$ hours         * Doubling Period ($d$) = $8$ hours     * Calculation: $N = 50,000(2)^{\frac{48}{8}} = 50,000(2)^6 = 50,000 \times 64 = 3,200,000$

• Continuous Compound Interest:     * Formula: $A = Pe^{rt}$     * Scenario: Investment of $\$2,000$ reaches $\$5,000$ after six years. Goal: Find interest rate.     * Variables:         * Principal ($P$) = $\$2000$         * Final Amount ($A$) = $\$5000$         * Time ($t$) = $6$     * Equation: $5,000 = 2,000e^{6r}$     * Solving for $r$: $2.5 = e^{6r} \implies \text{ln}(2.5) = 6r \implies r = \frac{\text{ln}(2.5)}{6}$

Expanding Logarithmic Expressions

• Key Expansion Rules:     * Product Rule: $\log_b(xy) = \log_b(x) + \log_b(y)$     * Quotient Rule: $\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)$     * Power Rule: $\log_b(x^k) = k \log_b(x)$

• Practice problems:     1. $\log_7(5xy) = \log_7(5) + \log_7(x) + \log_7(y)$     2. $\log(11m) = \log(11) + \log(m)$     3. $3\log(xyz) = 3(\log(x) + \log(y) + \log(z)) = 3\log(x) + 3\log(y) + 3\log(z)$     4. $\log_2(ab^3c^8) = \log_2(a) + 3\log_2(b) + 8\log_2(c)$     5. $\log(\frac{5x}{y^2z}) = \log(5) + \log(x) - (2\log(y) + \log(z)) = \log(5) + \log(x) - 2\log(y) - \log(z)$     6. $\log(\frac{ab^5}{x^2y^6z^9}) = \log(a) + 5\log(b) - (2\log(x) + 6\log(y) + 9\log(z)) = \log(a) + 5\log(b) - 2\log(x) - 6\log(y) - 9\log(z)$

Condensing Logarithmic Expressions

• Methodology: Reverse the expansion rules. Combined logs should share the same base.

• Practice problems:     1. $2\log(b) + \log(a) = \log(b^2) + \log(a) = \log(ab^2)$     2. $\log(a) - \log(b) = \log(\frac{a}{b})$     3. $\ln(5) + 2\ln(y) - \ln(4z) = \ln(5) + \ln(y^2) - \ln(4z) = \ln(\frac{5y^2}{4z})$     4. $3\ln(x) - 2\ln(y) - 3\ln(z) = \ln(x^3) - \ln(y^2) - \ln(z^3) = \ln(\frac{x^3}{y^2z^3})$     5. $3\log(a) + 3\log(b) - \log(c) = \log(a^3) + \log(b^3) - \log(c) = \log(\frac{a^3b^3}{c})$     6. $\log(x) + 3\log(y) - \log(z) = \log(x) + \log(y^3) - \log(z) = \log(\frac{xy^3}{z})$

Solving Logarithmic Functions

• Solution Strategies:     * If $\log_b(M) = \log_b(N)$, then $M = N$.     * If $\log_b(x) = y$, then convert to exponential form: $b^y = x$.     * Always condense multiple log terms on one side before solving.

• Specific Problems:     1. $\log_6(x + 3) + \log_6(4) = \log_6(14)$         * Condense: $\log_6(4(x + 3)) = \log_6(14)$.         * Equate Arguments: $4x + 12 = 14 \implies 4x = 2 \implies x = 0.5$.     2. $\log(x) + \log(8) = 2$         * Condense: $\log(8x) = 2$.         * Convert to Exponential (Common log base is $10$): $10^2 = 8x \implies 100 = 8x$.         * Solve: $x = 12.5$.     3. $\log_9(x + 6) - \log_9(3) = \log_9(2)$         * Condense: $\log_9(\frac{x + 6}{3}) = \log_9(2)$.         * Equate Arguments: $\frac{x + 6}{3} = 2 \implies x + 6 = 6 \implies x = 0$.     4. $\log_7(x) = 4\log_7(2) - \log_7(4)$         * Condense: $\log_7(x) = \log_7(2^4) - \log_7(4) = \log_7(\frac{16}{4})$.         * Equate Arguments: $x = 4$.     5. $2\log_7(2x) = 6$         * Isolate Log: $\log_7(2x) = 3$.         * Convert to Exponential: $7^3 = 2x \implies 343 = 2x$.         * Solve: $x = 171.5$.     6. $\log_3(x + 3) - 10 = -8$         * Isolate Log: $\log_3(x + 3) = 2$.         * Convert to Exponential: $3^2 = x + 3 \implies 9 = x + 3$.         * Solve: $x = 6$.", "title": "Study Guide: Solving Exponential and Logarithmic Functions and Application Problems" } @空json{