Problem 13 - Precalculus

Question

Solve the exponential equations exactly for \(x\). $$e^{5 x-1}=e^{x^{2}+3}$$

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = 4\).

Step 1: Set the Exponents Equal

Since the bases of the exponents are the same, we can set the exponents equal to each other. This gives us the equation:\[5x - 1 = x^2 + 3\]

Step 2: Rearrange the Equation

Rearrange the equation to form a quadratic equation. Move all terms to one side to get:\[x^2 - 5x + 4 = 0\]

Step 3: Factor the Quadratic Equation

Factor the quadratic equation. Look for two numbers that multiply to 4 and add to -5. The factors are -1 and -4. So, we can write:\[(x - 1)(x - 4) = 0\]

Step 4: Solve for x

Set each factor equal to zero and solve for \(x\):1. \(x - 1 = 0\) gives \(x = 1\).2. \(x - 4 = 0\) gives \(x = 4\).

Step 5: Verify the Solutions

Substitute \(x = 1\) and \(x = 4\) into the original expression to verify:1. For \(x = 1\), \(e^{5(1) - 1} = e^{4}\) and \(e^{1^2 + 3} = e^{4}\), which are equal.2. For \(x = 4\), \(e^{5(4) - 1} = e^{19}\) and \(e^{4^2 + 3} = e^{19}\), which are equal.Thus, both solutions are correct.

Key Concepts

Quadratic Equations

In mathematics, quadratic equations play a critical role. A quadratic equation is a type of polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable we're solving for. Quadratic equations are easily recognizable by the \(x^2\) term, and they can be found across various disciplines within mathematics. In our exercise, the quadratic equation formed is \(x^2 - 5x + 4 = 0\). This equation originated from the initial step of setting the exponents equal, giving a new focus for the problem.

Factoring

Factoring is a powerful method used to solve quadratic equations. It involves rewriting the quadratic equation in the form \((x - p)(x - q) = 0\), where \(p\) and \(q\) are the solutions. The challenge is to find numbers that multiply to give the constant term \(c\) and add up to give the middle term \(b\). For the equation \(x^2 - 5x + 4 = 0\), we sought two numbers that multiply to 4 (the constant term) and add to -5 (the middle term). These numbers are -1 and -4, enabling us to write the equation as \((x - 1)(x - 4) = 0\). Factoring transforms the quadratic into a simpler form that can be solved effortlessly.

Solving Equations

Once a quadratic equation is factored, solving it is straightforward. Each factor set to zero gives potential solutions. From the expression \((x - 1)(x - 4) = 0\), we find solutions by setting each parenthetical expression equal to zero.

• \(x - 1 = 0\) yields \(x = 1\).
• \(x - 4 = 0\) yields \(x = 4\).

These values of \(x\) are the potential solutions to the original exponential equation. Always verify that each solution is correct using the context of the original problem to ensure these values satisfy the given equation thoroughly.

Verification of Solutions

Verification of solutions is a crucial step in solving equations. After finding potential solutions, it is essential to ensure that they indeed satisfy the original equation. This involves substituting each solution back into the original equation to check for equality. For our solutions:

• When \(x = 1\), substituting into the original exponents \(e^{5x-1}\) and \(e^{x^2+3}\) gives \(e^4\) for both, confirming equality.
• Similarly, for \(x = 4\), substituting into both sides yields \(e^{19}\), verifying correctness.

This verification step is crucial as it confirms that both solutions \(x = 1\) and \(x = 4\) are valid and satisfy the given exponential equation, ensuring accuracy and completion of the solution process.