Advanced Mathematical Concepts: Comprehensive Notes on Natural Logarithms

Fundamentals and Definitions of Natural Logarithms Natural logarithms are specifically defined as logarithms that use the mathematical constant $e$ as their base. In mathematical notation, these are typically expressed as $\ln(x)$ rather than $\log_e(x)$. The transcript notes that any logarithm with a base other than $e$ can be successfully converted into a natural logarithm through the application of the change of base formula. Furthermore, the inverse operation of a natural logarithm is termed the antilogarithm, which is written as $\text{antiln}(x)$. These mathematical properties of logarithms and antilogarithms serve as foundational tools for simplifying and solving complex exponential and logarithmic equations, as well as inequalities. # The Change of Base Formula and Logarithmic Conversion The change of base formula is represented as $\log_a(n) = \frac{\log_b(n)}{\log_b(a)}$. To convert a generic logarithm into a natural logarithm, the base $b$ is substituted with $e$, resulting in the specific formula $\log_a(n) = \frac{\ln(n)}{\ln(a)}$. Example 1 in the study guide demonstrates this by converting $\log_4(381)$ to a natural logarithm. In this case, $a = 4$, $b = e$, and $n = 381$. The conversion yields the expression $\frac{\ln(381)}{\ln(4)}$. After utilizing a calculator to evaluate these natural logarithms, the result is approximately $4.2868$. Therefore, $\log_4(381) \approx 4.2868$. # Solving Equations Using Antilogarithms The antilogarithm is frequently employed to isolate and solve for variables within logarithmic equations. Example 2 provides the equation $3.75 = -7.5 \ln(x)$. To solve this, one must first divide both sides of the equation by $-7.5$, which simplifies the expression to $-0.5 = \ln(x)$. The next step involves taking the antilogarithm of each side: $\text{antiln}(-0.5) = x$. By inputting this into a calculator, the value for $x$ is determined to be approximately $0.6065$. # Step-by-Step Resolution of Exponential Equations with Natural Logarithms Natural logarithms are highly effective for solving exponential equations where bases are mismatched. In Example 3a, the equation $4^{3x} = 6^{x-1}$ is solved by taking the natural logarithm of both sides, resulting in $\ln(4^{3x}) = \ln(6^{x-1})$. Using the property $\ln(a^n) = n \ln(a)$, the equation is rewritten as $3x \ln(4) = (x-1) \ln(6)$. Substituting the numerical equivalents for the logarithms ($\ln(4) \approx 1.3863$ and $\ln(6) \approx 1.7918$), the equation becomes $3x(1.3863) = (x-1)(1.7918)$. This expands to $4.1589x = 1.7918x - 1.7918$. By subtracting $1.7918x$ from both sides, we get $2.3671x = -1.7918$, and dividing by $2.3671$ gives the solution $x \approx -0.7570$. (Note: while the transcript's final steps list the solution as $0.7570$, the algebraic steps imply the variable interaction with the minus sign in $(x-1)$). # Solving Inequalities with Natural Logarithms The study guide also addresses inequalities, such as 25 > e^{0.2t} in Example 3b. To solve for $t$, the natural logarithm is taken on both sides: \ln(25) > \ln(e^{0.2t}). Applying the power property of logarithms, this becomes \ln(25) > 0.2t \ln(e). Because $\ln(e)$ is equal to $1$, the inequality simplifies to \ln(25) > 0.2t. Using a calculator to find $\ln(25) \approx 3.2189$, the expression becomes 3.2189 > 0.2t. Dividing by $0.2$ results in 16.0945 > t, meaning that t < 16.0945. # Practical Application: Continuous Compounding in Banking A notable real-world application of natural logarithms is the calculation of interest that is compounded continuously. The formula for this process is $A = P e^{rt}$, where $A$ represents the total amount after time $t$, $P$ is the original principal sum, $r$ is the interest rate, and $t$ is the time in years. In a specific scenario, Ms. Cubbatz invested money on January 1, 1995, into a certificate of deposit with an $8\%$ ($0.08$) interest rate compounded continuously. By January 1, 1999, which represents a duration of $t = 4$ years, the account was worth $\$12,000$. To find the original investment, the formula is set up as $12,000 = P e^{0.08 \times 4}$. This requires solving for $P$ using natural logs and the property of base $e$. # Comprehensive Practice Problems and Evaluative Expressions The practice section includes sixteen problems designed to reinforce these concepts. Problems 1 through 3 require the evaluation of $\ln(71)$, $\ln(8.76)$, and $\ln(0.532)$. Problems 4 through 6 involve calculating antilogarithms: $\text{antiln}(-0.256)$, $\text{antiln}(4.62)$, and $\text{antiln}(-1.62)$. Problems 7 through 9 focus on converting logarithms to natural logarithms for evaluation, specifically $\log_7(94)$, $\log_5(256)$, and $\log_9(0.712)$. Finally, problems 10 through 15 present a variety of equations and inequalities to solve, including $6^x = 42$, $7^x = 4^{x-3}$, $1249 = 175 e^{-0.04t}$, $10^{x+1} = 3^x$, $12 = e^{0.048y}$, and $8.4 = e^{t-2}$. The final problem is the banking application for Ms. Cubbatz regarding her original principal investment.