Algebra and Trigonometry: Logarithmic and Exponential Equations

Section 6.6: Logarithmic and Exponential Equations

• This section covers the fundamental methodologies for solving equations involving logarithms and exponents, including algebraic manipulations and the use of graphing technology.

• Learning Objectives:     - Solving Logarithmic Equations.     - Solving Exponential Equations.     - Using a graphing utility to solve both logarithmic and exponential equations.

Fundamental Properties and Definitions

• Definition of a Logarithm:     - The relationship between logarithmic and exponential forms is defined as: $y = \log_{a}(x)$ if and only if $x = a^{y}$.     - Constraints for this definition: $M$, $N$, and $a$ must be positive, and $a \neq 1$.

• Logarithmic Property of Equality:     - If $M, N, \text{ and } a$ are positive and $a \neq 1$, then $\log_{a}(M) = \log_{a}(N)$ if and only if $M = N$.

• Exponential Property of Equality:     - If $a^{u} = a^{v}$, then $u = v$, provided that a > 0 and $a \neq 1$.

Solving Logarithmic Equations

• General Strategies:     - Identify the domain of the variable to avoid extraneous solutions.     - Use properties of logarithms (Product, Quotient, Power rules) to condense the equation into a single logarithm.     - Convert the logarithmic equation to its equivalent exponential form or use the property of equality.

• Example 1: Base 6 Logarithmic Equation:     - Equation: Solve $\log_{6}(x) + \log_{6}(x + 5) = 2$.     - Domain: The variable domain is x > 0.     - Step-by-Step:         1. Condense the logarithms: $\log_{6}(x(x + 5)) = 2$.         2. Convert to exponential form: $x(x + 5) = 6^{2}$.         3. Simplify: $x^{2} + 5x = 36$.         4. Standard quadratic form: $x^{2} + 5x - 36 = 0$.         5. Factor: $(x + 9)(x - 4) = 0$.         6. Solve: $x = -9$ or $x = 4$.     - Validation: Since the domain is x > 0, the value $-9$ is extraneous and must be discarded.     - Solution Set: ${4}$.

• Example 2: Base 5 Logarithmic Equation:     - Equation: Solve $\log_{5}(x + 9) + \log_{5}(x + 1) = 2$.     - Domain Requirements:         - x + 9 > 0 \rightarrow x > -9         - x + 1 > 0 \rightarrow x > -1         - Combined domain: x > -1.     - Procedure:         1. Express the left side as a single logarithm using the Product Property.         2. Change the equation to exponential form.     - Validation: Only $x = 1$ satisfies the restriction x > -1. The value $x = -11$ is extraneous.     - Solution Set: ${1}$.

• Example 3: Logarithmic Equality with Base 4:     - Equation: Solve $\log_{4}(x) + \log_{4}(x - 1) = \log_{4}(x + 15)$.     - Domain Requirements:         - x > 0.         - x - 1 > 0 \rightarrow x > 1.         - x + 15 > 0 \rightarrow x > -15.         - Combined domain: x > 1.     - Step-by-Step:         1. Use the log of a product property: $\log_{4}(x(x - 1)) = \log_{4}(x + 15)$.         2. Use the property of equality: $x(x - 1) = x + 15$.         3. Distribute: $x^{2} - x = x + 15$.         4. Standard form: $x^{2} - 2x - 15 = 0$.         5. Factor: $(x - 5)(x + 3) = 0$.     - Validation: Because the domain is x > 1, the value $-3$ is discarded as extraneous.     - Solution Set: ${5}$.

Solving Exponential Equations

• Example 4: Solving with Logs and Change of Base:     - Problem (a): Solve $4^{x} = 7$.         - Since $7$ cannot be written as an integer power of $4$, use the equivalent logarithmic form: $x = \log_{4}(7)$.         - Change of Base Formula: $x = \frac{\ln(7)}{\ln(4)}$.         - Alternative Method: Take the natural logarithm ($\ln$) or common logarithm ($\log$) of both sides: $\ln(4^{x}) = \ln(7)$ results in $x \ln(4) = \ln(7)$, so $x = \frac{\ln(7)}{\ln(4)}$.         - Approximate Solution: $1.404$.         - Solution Set: ${\frac{\ln(7)}{\ln(4)}\approx 1.404}$.     - Problem (b): Solve $e^{x} = 5$ (Implicitly referenced in slides as an exact match).         - Exact Solution: $x = \ln(5)$.         - Approximate Solution: $1.609$.

• Example 5: Bases that are not Powers of One Another:     - Equation: Solve $7^{x-1} = 2^{x}$.     - Strategy: Take the natural logarithm of both sides and apply the property $\ln(M^{r}) = r \ln(M)$.     - Step-by-Step:         1. $\ln(7^{x-1}) = \ln(2^{x})$.         2. $(x - 1) \ln(7) = x \ln(2)$.         3. Distribute: $x \ln(7) - \ln(7) = x \ln(2)$.         4. Collect terms with $x$: $x \ln(7) - x \ln(2) = \ln(7)$.         5. Factor $x$: $x(\ln(7) - \ln(2)) = \ln(7)$.         6. Solve for $x$: $x = \frac{\ln(7)}{\ln(7) - \ln(2)}$.     - Solution Set: ${\frac{\ln(7)}{\ln(7) - \ln(2)}}$.

Advanced Exponential Equations

• Example 6: Quadratic in Form:     - Condition: Recognizing that $4^{x} = (2^{2})^{x} = (2^{x})^{2}$.     - Equation Formulation: The original equation is manipulated into a quadratic form using substitution ($u = 2^{x}$).     - Solving Process:         - The substitution allows the equation to be written in a form such as $u^{2} - u - 12 = 0$ (logical progression from transcript context).         - The transcript notes: "The equation on the left has the solution $x = 1$, since $2^{1} = 2$."         - "The equation on the right has no solution, since 2^{x} > 0 for all $x$."     - Final Assessment: The only valid solution is $1$.     - Solution Set: ${1}$.

Graphical Solutions

• Example 7: Using a Graphing Utility:     - Problem: Solve $\ln(x) = -x$.     - Precision: Round solution to two decimal places.     - Methodology:         - Define $y_{1} = \ln(x)$ and $y_{2} = -x$.         - Observe that since $y_{1}$ is an increasing function and $y_{2}$ is a decreasing function, there is exactly one point of intersection.         - Utilize the INTERSECT command on the graphing utility.     - Result: The point of intersection reveals a solution of $x = 0.79$.