Problem 59 - Precalculus

Question

Solve the logarithmic equations. Round your answers to three decimal places. $$\ln \left(x^{2}+1\right)=4$$

Step-by-Step Solution

Answer

The solutions are \(x = 7.320\) and \(x = -7.320\).

Step 1: Understand the logarithmic equation

We are given the equation \(\left(x^{2}+1\right)=4\). This means that the natural logarithm of \(x^2+1\) is equal to 4.

Step 2: Remove the logarithm by exponentiation

To eliminate the natural logarithm, we exponentiate both sides of the equation. This involves using the property that \(e^{(a)} = a\). We write:\[(x^2+1)=e^4\]Which simplifies to:\[ x^2+1=e^4\].

Step 3: Solve for x^2

Now, isolate \(x^2\) by subtracting 1 from both sides of the equation:\(x^2 = e^4 - 1\).

Step 4: Calculate \(e^4\) and solve for x

Calculate \(e^4\) using a calculator, which is approximately 54.598. Substitute this back into the equation:\[x^2 = 54.598 - 1 = 53.598 \].Solve for \(x\) by finding the square root of both sides:\[x = \pm \sqrt{53.598}\].

Step 5: Compute the exact values of x

Calculate the square root of 53.598 using a calculator:\[\sqrt{53.598} \approx 7.320\].Thus, the solutions to the equation are \(x = 7.320\) and \(x = -7.320\).

Key Concepts

Natural Logarithm

Natural logarithms are fundamental in mathematics, especially when dealing with exponential growth or decay. The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718.

• Natural logarithms provide a way to reverse the process of exponentiation.
• They are useful in solving equations that involve exponential models.

In our original exercise, we worked with \( \ln(x^2 + 1) = 4 \). This equation implies that the power to which we raise \( e \) to get \( x^2 + 1 \) is 4. Hence, understanding natural logarithms is crucial for interpreting this expression. It allows us to think of the expression inside the logarithm as having been transformed by an exponentiation of base \( e \). When you exponentiate the other side, you simplify the equation by removing the \( \ln \).
This operation is equivalent to using the inverse function to isolate the value inside the logarithm. With this concept, you can better understand how logarithms are used to solve complex equations quickly and efficiently.

Exponentiation

Exponentiation involves raising a number to a power and is a critical concept in both algebra and logarithms. In the context of our original problem, using the property of exponentiation helped eliminate the natural logarithm.

• Exponentiation represents repeated multiplication.
• Essential for solving exponential equations and working with logarithms.

In solving the equation \( \ln(x^2 + 1) = 4 \), exponentiation is employed to convert \( \ln(x^2 + 1) \) into \( (x^2 + 1) = e^4 \).
This step is crucial since it simplifies the equation into a form that is easier to handle. The process involves recognizing that taking the exponent to the base \( e \) of \( \ln(x^2 + 1) \) gives \( x^2 + 1 \), thus, making it possible to proceed with algebraic manipulation. Exponentiation thus transforms logarithmic problems into solvable algebraic equations, a powerful step in mathematics.

Square Root Calculation

Calculating square roots is a key algebraic skill, especially in solving quadratic equations like the one derived in our exercise. After removing the logarithm and simplifying, the equation \( x^2 = 53.598 \) required solving for \( x \).

• The square root function finds the original value before it was squared.
• Used to solve equations of the form \( x^2 = a \).

To solve \( x^2 = 53.598 \), we take the square root of both sides, knowing that the operation will yield two possible values, one positive and one negative: \( x = \pm \sqrt{53.598} \).
This operation is essential because squaring is a non-reversible operation over the real numbers without considering the square's sign. Calculators can quickly find square roots to a decimal approximation, like how \( \sqrt{53.598} \approx 7.320 \). This precision reveals that both \( x = 7.320 \) and \( x = -7.320 \) are valid solutions to the equation, illustrating how square roots elucidate potential solution paths in algebra.