Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. e^2x - 6e^x + 8 = 0

Verified step by step guidance

1. Recognize that the equation involves exponential expressions with the same base \(e\). To simplify, let \(y = e^{x}\). Then, \(e^{2x} = (e^{x})^2 = y^2\).
2. Rewrite the original equation \(e^{2x} - 6e^{x} + 8 = 0\) in terms of \(y\) as \(y^2 - 6y + 8 = 0\).
3. Solve the quadratic equation \(y^2 - 6y + 8 = 0\) using factoring or the quadratic formula. This will give you the possible values for \(y\).
4. Recall that \(y = e^{x}\), so for each solution \(y_i\), solve the equation \(e^{x} = y_i\) by taking the natural logarithm of both sides: \(x = \ln(y_i)\).
5. Evaluate the logarithms to find the values of \(x\). Since the problem asks for decimal answers correct to the nearest thousandth, approximate the logarithms accordingly.

Key Concepts

Exponential Equations

Exponential equations involve variables in the exponent, such as e^x. Solving these requires understanding how to manipulate and isolate the exponential expression, often by substitution or applying logarithms.

Substitution Method for Quadratic Form

When an equation contains terms like e^{2x} and e^x, it can be treated as a quadratic by substituting u = e^x. This transforms the equation into a quadratic form, which can be solved using factoring or the quadratic formula.

Exact vs. Approximate Solutions

Exact solutions are expressed in terms of constants and variables without rounding, while approximate solutions are numerical values rounded to a specified decimal place. Understanding when to use each is important for interpreting answers correctly.