Problem 75 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the equation algebraically. Round your result to three decimal places. $$e^{2 x}+2 e^{x}-15=0$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 1.099\).

Step 1: Substitution

Let \(u = e^{x}\). Now, the equation becomes \(u^{2}+2u-15 = 0\).

Step 2: Solve the Quadratic Equation

Factor the quadratic equation: \((u-3)(u+5) = 0\). Set each factor equal to zero and solve for u: \(u-3=0\) resulting in \(u=3\), and \(u+5=0\) resulting in \(u=-5\).

Step 3: Substitute Back and Solve for x

Substitute \(u = e^{x}\) back into \(u=3\) to get \(e^{x}=3\). Solving for x gives \(x = \ln(3)\). Similarly, substituting \(u = e^{x}\) back into \(u=-5\) gives \(e^{x}=-5\), which is not possible since \(e^{x}\) can't be a negative number. Hence, discard this solution.

Step 4: Round off the Solution

Approximate the value of \(\ln(3)\) to three decimal places as 1.099.

Key Concepts

Algebraic Substitution

Algebraic substitution is a technique used to simplify and solve equations, especially when dealing with complex expressions. In the given problem, we encounter an exponential equation, which can be a bit intimidating at first.
To make this simpler, algebraic substitution comes into play. We choose a substitution like letting \( u = e^x \) which transforms the equation from an exponential form to a quadratic form, \( u^2 + 2u - 15 = 0 \).
This substitution helps in turning a complex problem into a more manageable problem by transforming it into a recognizable format that is easier to solve. By replacing the exponential expression \( e^x \) with \( u \), we can then solve it using methods suitable for quadratic equations. Once solved, we substitute back to find the actual solution for \( x \). This clever use of substitution is a powerful tool in algebra.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). They frequently appear in algebra and understanding them is crucial. In our problem, once the algebraic substitution is made, the equation transforms into \( u^2 + 2u - 15 = 0 \).
This is now a typical quadratic equation. To solve it, one can either factor the quadratic or use the quadratic formula. In this case, the equation is factorable: \((u-3)(u+5) = 0\).

• Setting \( u - 3 = 0 \), we solve for \( u \) and find \( u = 3 \).
• Setting \( u + 5 = 0 \), we solve for \( u \) and find \( u = -5 \), but this is not usable since \( u = e^x \) must be positive.

Factoring is a quick and efficient way to find solutions, as long as the equation can be neatly factored. Recognizing factorable quadratics helps in swiftly solving problems like this one.

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are logarithms with base \( e \), where \( e \) is approximately 2.71828. They are widely used in solving problems involving exponentials.
In our specific problem, after making the substitutions and solving the quadratic equation, we are left with one potential solution: \( e^x = 3 \).
Here, we use natural logarithms to solve for \( x \) by considering the relationship \( x = \ln(3) \).

• The natural logarithm function is the inverse of the exponential function, which makes it ideal for solving equations where the variable is an exponent.
• Next, calculate \( \ln(3) \) to find the value of \( x \).
• Using a calculator, \( \ln(3) \) is approximately 1.099.

Thus, natural logarithms help in unlocking the solutions to exponential equations by providing a straightforward way to isolate and solve for the variable.