Problem 60 - College Algebra with Modeling and Visualization

Question

Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. $$e^{-x}=\frac{1}{2}$$

Step-by-Step Solution

Answer

The solution is \(x \approx 0.69\).

Step 1: Isolate the Exponential Expression

Identify the exponential expression in the equation and ensure it is isolated. Here, the equation is already in the form where the exponential expression \(e^{-x}\) is isolated.

Step 2: Apply the Natural Logarithm

To solve for \(x\), take the natural logarithm of both sides of the equation. This gives:\[ \ln(e^{-x}) = \ln\left(\frac{1}{2}\right) \] The natural log helps us bring the exponent down.

Step 3: Simplify the Logarithmic Equation

Using the logarithmic identity \( \ln(e^a) = a \), simplify the left side of the equation:\[ -x = \ln\left(\frac{1}{2}\right) \]

Step 4: Solve for x

To solve for \(x\), multiply both sides by -1:\[ x = -\ln\left(\frac{1}{2}\right) \]

Step 5: Change of Base Formula

Since \(\ln\) is being used, it's unnecessary to apply the change of base. However, you can employ it if you'd use common logarithms in other contexts. Here:\[ x \approx -\ln(0.5) \]

Step 6: Approximate Solution

Calculate \( \ln(0.5) \) using a calculator and round to the nearest hundredth:\[ \ln(0.5) \approx -0.693 \] Therefore, \[ x \approx -(-0.693) = 0.693 \]

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a special type of logarithm. It uses the mathematical constant \(e\) as its base. The constant \(e\) is approximately equal to 2.71828 and is a fundamental constant in mathematics, especially in calculus and complex analysis.

The natural logarithm is predominantly used because it simplifies many mathematical equations, especially those involving exponential growth or decay. The primary property of a natural logarithm is that it allows us to "bring down" exponents in equations.

• For instance, if you have \(e^x\) and you want to solve for \(x\), taking the natural log of both sides \(\ln(e^x) = x\) simplifies the process vastly.
• Thus, the natural logarithm transforms a multiplication problem (exponent) into an addition problem, making it easier to work with.

In the exercise, taking the natural logarithm of both sides \(\ln(e^{-x}) = \ln(0.5)\) allowed us to bring down the exponent \(-x\) and solve for \(x\), thus showing its immense utility in solving exponential equations.

Change of Base Formula

The change of base formula is very useful when you're dealing with a logarithm that isn't in the base you'd like to work with, or when you don't have a calculator with the right button. The formula is:

• \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\)

This allows you to convert any logarithm to a base you are more comfortable with, usually base 10 (\(\log_{10}\)) or base \(e\) (\(\ln\)).

In the given problem though, the base \(e\) is already the natural choice. Hence, direct application of the natural log was most efficient. However, if you had to use another base, say 10, you'd then employ the change of base to find the solution. In practical terms, if you needed \(\log_{10}(0.5)\), and were only familiar with natural logs, you could convert it with this concept:

• \(\log_{10}(0.5) = \frac{\ln(0.5)}{\ln(10)}\)

Knowing the change of base formula enhances versatility in solving logarithmic equations with whatever resources you have at hand, like various calculators.

Logarithmic Identity

Logarithmic identities are helpful tools when simplifying equations. They represent basic truths about logarithms that can be used to make calculations easier. Among these identities, the one often used in conjunction with natural logarithms is:

• \(\ln(e^a) = a\)

This identity states that the natural logarithm of \(e\) raised to any power \(a\) is simply \(a\). It is derived from the definition of logarithms as the inverse functions of exponentials.

In the solved exercise, the logarithmic identity \(\ln(e^{-x}) = -x\) was pivotal in simplifying the equation. It eliminated the complex factor of having an exponent by making it linear, showing just how powerful these fundamental properties can be. Being familiar with logarithmic identities enhances one's ability to work through exponential equations and further understand their behavior.