Solve each exponential equation in Exercises 23–48. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. e^2x−3e^x+2=0

Verified step by step guidance

1. Recognize that the equation involves exponential expressions with base e: $e^{2x}-\; 3e^x+\; 2\; =\; 0$.
2. Make a substitution to simplify the equation: let $t\; =\; e^x$. Since $e^\{2x\}\; =\; (e^x)^2$, rewrite the equation as $t^2\; -\; 3t\; +\; 2\; =\; 0$.
3. Solve the quadratic equation $t^2\; -\; 3t\; +\; 2\; =\; 0$ by factoring or using the quadratic formula. The factored form is $(t\; -\; 1)(t\; -\; 2)\; =\; 0$, so the solutions for t are $t\; =\; 1$ and $t\; =\; 2$.
4. Recall the substitution $t\; =\; e^x$. Set each solution equal to $e^x$: $e^x\; =\; 1$ and $e^x\; =\; 2$. Solve for x by taking the natural logarithm of both sides: $x\; =\; \backslash ,\; \backslash ln(1)$ and $x\; =\; \backslash ,\; \backslash ln(2)$.
5. Express the solution set as $\backslash \{\; x\; =\; \backslash ln(1),\; x\; =\; \backslash ln(2)\; \backslash \}$. Since $\backslash ln(1)\; =\; 0$, the solutions are $x\; =\; 0$ and $x\; =\; \backslash ln(2)$. Use a calculator to approximate $\backslash ln(2)$ to two decimal places.

Key Concepts

Exponential Equations

Exponential equations involve variables in the exponent, such as e^(2x) or e^x. Solving these requires rewriting the equation to isolate the exponential term or transforming it into a quadratic form by substitution, enabling the use of algebraic methods.

Logarithms and Their Properties

Logarithms are the inverse operations of exponentials, allowing us to solve for variables in exponents. Natural logarithms (ln) use base e, while common logarithms use base 10. Applying logarithms helps convert exponential equations into linear ones for easier solving.

Quadratic Equation Techniques

When an exponential equation can be rewritten as a quadratic (e.g., by substituting u = e^x), standard methods like factoring, completing the square, or the quadratic formula can be used. This approach simplifies solving for the substituted variable before back-substituting.