Problem 47 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ (\ln x)^{2}+\ln x=2 $$

Step-by-Step Solution

Answer

The solutions are \(x = e^{-2}\) and \(x = e\).

Step 1: Rewrite the Equation

First, let's set up the equation by substituting \( y = \ln x \). So the equation \((\ln x)^2 + \ln x = 2\) becomes \(y^2 + y - 2 = 0\).

Step 2: Identify a Suitable Method

The given equation \(y^2 + y - 2 = 0\) is a quadratic equation. We can solve this by factoring because it is in the form \(ax^2 + bx + c = 0\).

Step 3: Factor the Quadratic Equation

To factor \(y^2 + y - 2 = 0\), find two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\). So, factor the equation as \((y+2)(y-1) = 0\).

Step 4: Solve for y

Set each factor to zero: 1. \(y + 2 = 0\) gives \(y = -2\).2. \(y - 1 = 0\) gives \(y = 1\).

Step 5: Solve for x

Substitute back \(y = \ln x\):1. For \(y = -2\), we have \(\ln x = -2\). Solve this by exponentiating both sides: \(x = e^{-2}\).2. For \(y = 1\), we have \(\ln x = 1\). Solve this by exponentiating both sides: \(x = e^{1} = e\).

Step 6: Verify the Solutions

Verify the calculated values of \(x\):- For \(x = e^{-2}\), substitute back into the original equation: \((\ln(e^{-2}))^2 + \ln(e^{-2}) = 2\) simplifies to \(4 + (-2) = 2\).- For \(x = e\), substitute back into the original equation: \((\ln(e))^2 + \ln(e) = 2\) simplifies to \(1 + 1 = 2\). Both satisfy the original equation.

Key Concepts

Factoring

Factoring is a key technique for solving quadratic equations like the one in the exercise. It involves expressing the quadratic equation in the form of a product of two binomials. In our example, the transformed equation is \(y^2 + y - 2 = 0\). To factor this, you need to find two numbers that multiply to \(-2\) (the constant term \(c\)) and add to \(1\) (the coefficient of \(y\)).

In this case, the numbers are \(2\) and \(-1\), and thus, the equation factors to \((y+2)(y-1) = 0\). Once we have the equation in factored form, solving for \(y\) becomes straightforward: set each factor to zero and solve for \(y\). This yields the solutions \(y = -2\) and \(y = 1\), which are pivotal for the next steps of solving for \(x\).

This method is efficient and particularly useful when the quadratic equation can easily be factored, saving the need for more complex solutions like the quadratic formula. Remember, successful factoring requires practice in recognizing number patterns and relationships.

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This method is essential, especially when factoring is not straightforward or possible. The beauty of the quadratic formula is its ability to yield solutions even when the factors of the equation are not integers.

When using the quadratic formula, always calculate the discriminant (the part under the square root, \(b^2 - 4ac\)).

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If it is negative, the solutions are complex numbers.

While factoring wasn't necessary in our primary equation, understanding how to apply the quadratic formula is crucial for tackling more complex equations that don't simplify as easily. It's a fail-safe method where factoring might falter.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and play a vital role when dealing with equations involving variables within a logarithm. In this exercise, we encounter the natural logarithm, denoted as \(\ln\), which specifically uses base \(e\).

To solve the equation \((\ln x)^2 + \ln x = 2\), we substitute \(y = \ln x\) to transform the equation into a more familiar quadratic form. Once solved for \(y\), we back-substitute to find \(x\). Because \(y = \ln x\), the solutions \(y = -2\) and \(y = 1\) translate into \(x = e^{-2}\) and \(x = e^1 = e\) respectively by exponentiating.

It's important to understand that logarithmic functions can simplify seemingly complex problems by converting exponential relationships into linear ones, making them manageable and easier to handle in the context of solving equations.