Problem 578 - Precalculus

Question

Find the exact solution for \(e^{2 x}-e^{x}-72=0 .\) If there is no solution, write no solution.

Step-by-Step Solution

Answer

The solution for the equation is \(x = \ln(9)\).

Step 1: Simplify the Expression

The given equation is \(e^{2x} - e^x - 72 = 0\). Let's substitute \(y = e^x\). Then \(e^{2x} = (e^x)^2 = y^2\). This substitution transforms the equation into a quadratic equation: \(y^2 - y - 72 = 0\).

Step 2: Factor the Quadratic

The equation \(y^2 - y - 72 = 0\) can be factored by finding two numbers that multiply to \(-72\) and add to \(-1\). These numbers are \(-9\) and \(8\). Thus, we can factor the equation as: \((y - 9)(y + 8) = 0\).

Step 3: Solve for y

Set each factor equal to zero. For \(y - 9 = 0\), we get \(y = 9\). For \(y + 8 = 0\), we get \(y = -8\). Since \(y = e^x\) and \(e^x\) must be positive, \(y = -8\) is not applicable. So, \(y = 9\) is the only solution.

Step 4: Solve for x

Since we have \(y = e^x = 9\), we can solve for \(x\) by taking the natural logarithm of both sides: \(x = \ln(9)\).

Key Concepts

Substitution Method

To solve exponential equations like the one given in the exercise, a useful approach is the substitution method. This method helps simplify complex exponential expressions by introducing a new variable for the exponent.

In our exercise, we initially have an exponential equation: \(e^{2x} - e^x - 72 = 0\). This can look intimidating at first glance. By using substitution, we make things more manageable.

We set \(y = e^x\). Substituting this into the equation gives us \(e^{2x} = y^2\), which turns the original equation into a quadratic: \(y^2 - y - 72 = 0\).

The substitution method is a powerful tool in mathematics. It not only simplifies the structure of the problem but also frequently bridges gaps between different types of mathematical expressions, making them easier to handle.

Quadratic Equations

Quadratic equations like the transformed form \(y^2 - y - 72 = 0\) are second-degree polynomials. These can usually be solved by factoring, completing the square, or using the quadratic formula.

In this problem, we can factor the quadratic equation. We need to find two numbers that multiply to \(-72\) and sum to \(-1\). These numbers are \(-9\) and \(8\), so we can express the equation as \((y - 9)(y + 8) = 0\).

By setting each factor equal to zero, we find the potential solutions for \(y\). Thus, \(y = 9\) and \(y = -8\). However, since \(y\) represents \(e^x\), which must be positive, only \(y = 9\) is acceptable.

Quadratic equations are foundational in algebra and appear in various contexts beyond this one, making them essential knowledge for students.

Natural Logarithm

After determining that \(y = 9\) is the valid solution from the quadratic equation, we return to our substitution \(y = e^x\). We solve for \(x\) by using logarithms, specifically the natural logarithm.

The natural logarithm, denoted as \(\ln\), is the inverse of the exponential function \(e^x\). So, to find \(x\), we take the natural logarithm of both sides: \(x = \ln(9)\).

This step highlights the usefulness of \(\ln\) in solving exponential equations.

• It is particularly crucial in calculus and in solving equations where the variable is an exponent.
• Understanding how to manipulate exponential and logarithmic functions is key to advancing in mathematics.