Problem 73 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the indicated variable. $$A=T_{0}+C e^{-k}, \text { for } k$$

Step-by-Step Solution

Answer

Solve for \( k \) with \( k = -\ln \left( \frac{A - T_{0}}{C} \right) \).

Step 1: Isolate the Exponential Factor

Start by isolating the exponential term that contains the variable of interest, which is the expression with the variable \( k \). The given formula is \( A = T_{0} + C e^{-k} \). To begin, subtract \( T_{0} \) from both sides to get the equation \( A - T_{0} = C e^{-k} \).

Step 2: Solve for the Exponential Term

Next, divide both sides by \( C \) to solve for the exponential, giving \( \frac{A - T_{0}}{C} = e^{-k} \). This expression now isolates the exponential function.

Step 3: Apply the Natural Logarithm

To isolate \( k \), apply the natural logarithm (\( \ln \)) to both sides of the equation. This gives \( \ln \left( \frac{A - T_{0}}{C} \right) = -k \), because the natural log of an exponent nullifies it according to the logarithmic property: \( \ln(e^x) = x \).

Step 4: Solve for Variable k

The final step is to solve for \( k \) by multiplying both sides by \(-1\). This results in \( k = -\ln \left( \frac{A - T_{0}}{C} \right) \). Now, \( k \) is isolated and solved in terms of \( A \), \( T_{0} \), and \( C \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where the variables appear as exponents. They are often used to model growth and decay phenomena, such as population growth or radioactive decay. In the context of the given problem, the term \( C e^{-k} \) represents an exponential function. Here, \( C \) is the coefficient, and \(-k\) is the exponent. The base of the exponential function is \( e \), Euler's number, approximately equal to 2.718. This particular type of function is common in natural processes and appears in equations modelling naturally occurring phenomena.

Exponential functions have some unique properties. For example, the value of the exponential function becomes very small as \( k \) increases, due to the negative exponent. They also tend to behave predictably, making them useful for extrapolating data over time periods. Understanding how to manipulate these functions is crucial for solving equations like the one in the exercise.

Natural Logarithms

Natural logarithms are logarithms that have Euler's number \( e \) as their base. Indicated by \( \ln \), they are the inverse operations of exponential functions. This means if \( e^x = y \), then \( \ln(y) = x \). In the problem, applying a natural logarithm helps in isolating the variable \( k \) from the exponential function \( e^{-k} \).

Taking the natural logarithm of both sides of an equation where the variable is an exponent allows you to simplify the expression significantly. This is because the natural logarithm has the special property of neutralizing the exponential function: \( \ln(e^x) = x \). This property is key for solving equations involving exponential functions since it allows you to handle them in a more algebraic manner. Using natural logarithms is a vital technique in calculus and algebra for solving exponential equations.

Variable Isolation

Variable isolation is the process of rearranging an equation to make one variable the subject, or to solve for one specific variable. In the context of our problem, we are tasked to isolate \( k \). This is done step-by-step by using mathematical operations that maintain the equality of the equation.

The procedure starts by isolating the term that includes the variable \( k \) from other terms. In our example, we first subtract \( T_{0} \) from both sides of the equation to remove it from the right side. Next, divide the resulting expression by \( C \) to truly isolate the exponential term. To ultimately isolate \( k \), we take the natural logarithm of both sides and then multiply by \(-1\) to eliminate the negative sign.

Isolating a variable is an essential skill in algebra and helps to solve for unknowns, making it possible to understand relationships between different quantities in an equation. Mastering this skill facilitates problem-solving across various fields of mathematics and science.