Problem 530 - Precalculus

Question

Find the exact solution for \(e^{2 x}-e^{x}-110=0 .\) If there is no solution, write no solution.

Step-by-Step Solution

Answer

The solution is \( x = \ln(11) \).

Step 1: Substitute Variable

Let's set a substitution to simplify the equation. Let \( y = e^x \). So, \( e^{2x} = (e^x)^2 = y^2 \). The equation now becomes \( y^2 - y - 110 = 0 \).

Step 2: Solve the Quadratic Equation

We now solve the quadratic equation \( y^2 - y - 110 = 0 \) using the quadratic formula. The quadratic formula is given by \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -110 \).

Step 3: Apply Quadratic Formula

Plug the values into the quadratic formula:\[y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-110)}}{2(1)}\]\[y = \frac{1 \pm \sqrt{1 + 440}}{2}\]\[y = \frac{1 \pm \sqrt{441}}{2}\]\[y = \frac{1 \pm 21}{2}\]

Step 4: Determine Values for y

Solving \( y = \frac{1 + 21}{2} \) gives \( y = 11 \). Solving \( y = \frac{1 - 21}{2} \) gives \( y = -10 \). Since \( y = e^x \) must be positive, we only consider \( y = 11 \).

Step 5: Solve for x

Since \( y = e^x = 11 \), we take the natural logarithm of both sides to solve for \( x \):\[ x = \ln(11) \].

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They commonly take the form \(f(x) = a \cdot b^x\), where \(a\) is a constant, \(b\) is the base, and \(x\) is the exponent. In our exercise, the function \(e^{2x} - e^x - 110 = 0\) includes the transcendental number \(e\) as the base, which is approximately 2.718 and is frequently used in mathematics. Exponential functions have some key properties:

• They are always positive for any real number exponent when the base is positive.
• They have a constant rate of growth or decay, which results in their characteristic curved graphs.
• They are essential in modeling natural processes, like population growth and radioactive decay, as they represent continuously compounded growth.

In the solution, we start with an expression involving \(e^{2x}\) and \(e^x\) and simplify it using effective substitution, making it easier to solve the equation by transforming it into a familiar quadratic form.

Natural Logarithms

The natural logarithm, represented by \(\ln\), is the inverse function of the exponential function with base \(e\). It allows us to solve equations involving exponents. The natural logarithm gives you the power to which \(e\) must be raised to obtain a given number. For example, if \(y = e^x\), then \(x = \ln(y)\).

• Natural logarithms can solve equations with exponents by transforming multiplicative relationships into additive ones.
• They are used in a variety of disciplines such as biology, economics, and engineering, often in equations that model natural growth or decay.
• Understanding \(\ln(x)\) is crucial for moving from exponential problems to linear ones, making equations easier to handle mathematically.

In our exercise, after determining that \(y = 11\), we can solve for \(x\) with \(x = \ln(11)\). This technique is effective in isolating the variable \(x\) when initially embedded within an exponent.

Solving Equations

Solving equations involves finding the values for variables that satisfy the equation. This process often requires several steps to simplify the equation and find a solution. Here's how we approached solving the given problem:

• First, a substitution was made to transform the given exponential equation into a quadratic equation. This simplified the complex expression and made it more manageable.
• The quadratic equation \(y^2 - y - 110 = 0\) was then solved using the quadratic formula, which provides solutions for equations of the form \(ax^2 + bx + c = 0\).
• Key to solving the quadratic equation is determining the discriminant \(b^2 - 4ac\), which influences the number and nature of the solutions. In this case, the discriminant was a perfect square, leading to two real solutions for \(y\).
• We verified solutions by checking that \(y > 0\) since \(y = e^x\) must be positive, and then calculated \(x = \ln(11)\) for the valid solution.

By systematically applying substitution, the quadratic formula, and natural logarithms, we effectively simplified and solved the equation.