Problem 17 - College Algebra

Question

For the following exercises, use logarithms to solve. $$ e^{-3 k}+6=44 $$

Step-by-Step Solution

Answer

The value of \( k \) is approximately -1.2125.

Step 1: Isolate the Exponential Expression

Start by subtracting 6 from both sides of the equation.\[ e^{-3k} + 6 = 44 \]Subtract 6:\[ e^{-3k} = 44 - 6 \]\[ e^{-3k} = 38 \]

Step 2: Apply the Natural Logarithm

Take the natural logarithm of both sides to eliminate the exponential function.\[ \ln(e^{-3k}) = \ln(38) \]Using the property of logarithms \( \ln(a^b) = b\ln(a) \), this simplifies to:\[ -3k \cdot \ln(e) = \ln(38) \]Since \( \ln(e) = 1 \), we have:\[ -3k = \ln(38) \]

Step 3: Solve for k

Divide both sides by -3 to solve for \( k \).\[ k = \frac{\ln(38)}{-3} \]

Step 4: Calculate the Value

Use a calculator to find the value of \( \ln(38) \) and then divide by -3.\[ \ln(38) \approx 3.6376 \]\[ k \approx \frac{3.6376}{-3} \approx -1.2125 \]

Key Concepts

Exponential Equations

Exponential equations are equations where variables appear in exponents. They often have the form \( a^x = b \) where \( x \) is the exponent we are trying to solve. To find the value of the unknown variable in such equations, we often use logarithms because they are the inverse operations of exponential functions.

Some basic characteristics of exponential equations include:

• They express changes that occur proportionally over time, such as growth or decay.
• Exponential growth sees values increasing rapidly, such as population growth, while decay models things like radioactive decay.

When solving exponential equations, the key is to transform them into a linear form that can easily be solved. This is usually done by isolating the exponential expression before applying a logarithm.

Natural Logarithm

A natural logarithm is a logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. It is denoted as \( \ln \), as in \( \ln(x) \). The natural logarithm is especially useful in dealing with exponential equations involving base \( e \).

Key properties of natural logarithms include:

• The natural logarithm of \( e \) itself is 1, i.e., \( \ln(e) = 1 \).
• Using natural logarithms can simplify problems, especially when the equations involve \( e \).
• Natural logarithms have numerous applications in calculus, particularly integration and differentiation.

To use natural logarithms in solving the exercise, we took the logarithm on both sides of the isolated exponential equation. This detached the exponent \(-3k\) from the base \( e \), transforming the equation into a form we could algebraically solve.

Isolation of Variables

Isolation of variables is a process used in algebra to rearrange equations so that you solve for one specific variable. It involves manipulating the equation to get the variable of interest by itself on one side.

Here's how variable isolation works in our step-by-step solution:

• First, the exponential expression was isolated by removing constants from the equation's left side.
• By subtracting 6 from both sides, we simplified it to only have the exponential term \( e^{-3k} \) on one side.
• Once isolated, we were able to apply logarithms, which let us further isolate \( k \).

Isolation allows us to methodically solve equations, making complex problems more manageable by breaking them down into simpler parts. This strategic approach is foundational in algebra and calculus, as it ensures clarity and precision when finding solutions.