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^4x+5e^2x−24=0

Verified step by step guidance

1. Start by recognizing that the equation involves exponential expressions with different exponents: \(e^{4x} + 5e^{2x} - 24 = 0\). Notice that \(e^{4x}\) can be rewritten as \((e^{2x})^2\) to simplify the equation.
2. Make a substitution to turn the equation into a quadratic form. Let \(u = e^{2x}\). Then the equation becomes \(u^2 + 5u - 24 = 0\).
3. Solve the quadratic equation \(u^2 + 5u - 24 = 0\) using the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=5\), and \(c=-24\).
4. After finding the values of \(u\), substitute back \(u = e^{2x}\) and solve for \(x\) by taking the natural logarithm: \$2x = \ln(u)\(, so \)x = \frac{1}{2} \ln(u)$.
5. Evaluate the logarithmic expressions using a calculator to find decimal approximations of \(x\), rounding to two decimal places as required.

Key Concepts

Exponential Equations

Exponential equations involve variables in the exponent, such as e^(4x). Solving these requires rewriting the equation to isolate the exponential term or transform it into a quadratic form, enabling the use of algebraic methods to find the variable.

Substitution Method

When an exponential equation contains terms like e^(4x) and e^(2x), substitution can simplify it. For example, letting u = e^(2x) transforms the equation into a quadratic in u, which can be solved using factoring or the quadratic formula.

Logarithms and Their Properties

Logarithms are the inverse of exponentials and are used to solve for variables in exponents. After isolating the exponential expression, applying natural logarithms (ln) or common logarithms (log) helps find the exact solution, which can then be approximated with a calculator.