Problem 76 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the equation algebraically. Round your result to three decimal places. $$e^{2 x}-10 e^{x}+24=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 1.792\) and \(x = 1.386\), rounded to three decimal places.

Step 1: Recognize the structure and make an appropriate substitution

The equation \(e^{2x} - 10e^{x} + 24 = 0\) looks like a quadratic equation in the form \(u^2 - 10u + 24 = 0\) if we let \(u = e^{x}\). So let's substitute \(u = e^{x}\). The equation becomes \(u^2 - 10u + 24 = 0\).

Step 2: Solve the quadratic equation

The quadratic equation \(u^2 - 10u + 24 = 0\) can be solved using the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\). For this equation, \(a = 1\), \(b = -10\) and \(c = 24\). Substituting these values into the formula gives two values for \(u\): \(u_1 = \frac{10 - \sqrt{100 - 96}}{2} = 6\), \(u_2 = \frac{10 + \sqrt{100 - 96}}{2} = 4\).

Step 3: Substitute \(u\) back

Substitute \(u = e^{x}\) back into \(u_1\) and \(u_2\), we get \(e^{x} = 6\) and \(e^{x} = 4\). Taking natural logarithms on each side of these two equations, we obtain \(x = ln(6)\) and \(x = ln(4)\).

Step 4: Calculate the numerical values

Now we can calculate the numerical values of \(x\). We have \(x_1 = ln(6) = 1.792\) and \(x_2 = ln(4) = 1.386\), rounded to three decimal places.

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra and appear in various forms across different problems. A quadratic equation is typically in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. The solutions to these equations are found using different methods such as factoring, completing the square, or the quadratic formula.
When an equation mimics this form but involves higher powers or different functions, such as in exponential forms, substitution methods often bring it into a quadratic format.
Key points:

• Quadratic equations have two solutions, which may be real or complex.
• The solutions can be determined using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
• If the discriminant \(b^2 - 4ac\) is positive, there are two distinct real solutions; if zero, one real solution; and if negative, two complex solutions.

Understanding quadratic equations and how to solve them is crucial for tackling more complex algebraic equations.

Substitution Method

The substitution method is a technique used to simplify complex equations, particularly when transforming non-polynomial equations into quadratic ones. It's especially useful when dealing with equations where traditional algebraic manipulation is challenging.
In the given problem, we utilized substitution by recognizing the structure that resembled a quadratic equation. Here’s how it works:

• Identify a substitution that can simplify the equation. In this scenario, we substituted \(u = e^{x}\), transforming the equation from exponential form to quadratic form: \(u^2 - 10u + 24 = 0\).
• Once substituted, solve the transformed equation using standard techniques like the quadratic formula or factoring.
• Substitute back the original expression to get the final solutions in terms of the original variables. This involves expressing potential solutions like \(e^{x} = 6\) and \(e^{x} = 4\), and finding \(x\) using natural logarithms.

Using substitution makes it feasible to tackle problems that would otherwise require complicated computations or techniques.

Natural Logarithm

Natural logarithms are vital in solving equations involving exponential functions. The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is approximately equal to 2.71828.
They simplify finding the values of variables involved in exponential equations, as the natural log of \(e^{x}\) simplifies directly to \(x\).
In the problem at hand:

• After finding solutions \(e^{x} = 6\) and \(e^{x} = 4\), the natural logarithm is used: \(x = \ln(6)\) and \(x = \ln(4)\).
• The natural log effectively allows us to solve exponential equations. Since the function \(\ln\) is a logarithmic inverse of the exponential function, these solutions provide the specific \(x\) values that satisfy the original exponential equation.
• Calculating the actual values of \(x\) results in precise numerical examples: \(x = 1.792\) and \(x = 1.386\), rounded to three decimal places.

Understanding natural logarithms is crucial for performing operations that involve the base \(e\) and interpreting equations with exponential growth or decay.