Problem 113 - A Graphical Approach to College Algebra

Question

Use any method (analytic or graphical) to solve each equation. $$\ln \left(\ln e^{-x}\right)=\ln 3$$

Step-by-Step Solution

Answer

The solution is \(x = -3\).

Step 1: Understanding the Equation

The given equation is \(\ln (\ln e^{-x}) = \ln 3\). We need to find the value of \(x\) that satisfies this equation.

Step 2: Simplify the Inside Expression

The inner expression is \(\ln e^{-x}\). Use the property \(\ln(e^a) = a\), which gives: \[\ln e^{-x} = -x\]

Step 3: Substitute and Simplify

Replacing \(\ln e^{-x}\) with \(-x\) in the original equation gives: \[\ln (-x) = \ln 3\]

Step 4: Solve for x

Since the logs on both sides are equal, set the arguments equal to each other:\[-x = 3\]Solve for \(x\):\[x = -3\]

Step 5: Verify the Solution

Substitute \(x = -3\) back into the original equation to ensure it holds true:\[\ln (\ln e^{-(-3)}) = \ln (\ln e^3) = \ln 3\]Since both sides are equal, \(x = -3\) is the correct solution.

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms of variables or constants. Solving them usually requires us to apply the properties of logarithms to manipulate the equation into a solvable form. Logarithmic equations often appear in different formats, like natural logarithms (\(\ln\)) or common logarithms (\(\log_{10}\)).

When encountering a logarithmic equation like \(\ln (\ln e^{-x}) = \ln 3\), our goal is to isolate the variable by equating the arguments of the logarithms on both sides. The logarithmic identity \(\ln(a) = \ln(b)\) implies that \(a = b\). This principle greatly simplifies the task of solving logarithmic equations by turning them into algebraic equations.

Properties of Logarithms

To solve logarithmic equations, understanding the properties of logarithms is crucial. Some useful properties include:

• \(\ln(ab) = \ln a + \ln b\)
• \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\)
• \(\ln(a^b) = b\ln a\)
• \(\ln(e^a) = a\) (natural exponential and logarithm are inverses)

These properties help simplify expressions and solve equations more efficiently. For example, in the given equation \(\ln e^{-x} = -x\), we used the property \(\ln(e^a) = a\) to simplify inside expressions and reduce complexity.

Mastery of these properties aids in recognizing equivalent expressions and provides shortcuts for more complex problems.

Solving Equations Analytically

Solving equations analytically involves manipulating the equation algebraically to find a solution. In our example, we initially had the equation \(\ln (\ln e^{-x}) = \ln 3\).

We began by simplifying the inner expression \(\ln e^{-x}\) by applying logarithmic properties, yielding \(-x\). Substituting this back into the equation resulted in \(\ln(-x) = \ln 3\).

The next step was to use the property that if two logarithms with the same base are equal, then their arguments are equal: \(-x = 3\).
Solving for \(x\) gives us \(x = -3\). Overlaying algebraic manipulation on top of logarithmic identities allows us to transform complex expressions into solvable algebraic forms.

Verification of Solutions

After finding a solution, it is essential to verify that the solution is correct by substituting it back into the original equation. Verification ensures that no mistakes were made during the manipulation process. For our equation, substituting \(x = -3\) back gives:

\[\ln(\ln e^{-(-3)}) = \ln(\ln e^3) = \ln 3\]
Since both sides of the equation match, it confirms that \(x = -3\) is indeed correct.

Verification not only confirms accuracy but also boosts confidence in the method used. It highlights the importance of retracing steps and validates that all mathematical operations were properly applied, leading to a trustworthy conclusion.