Question

solve the following exponential equation. express the solution set in terms of natural logarithms or common logarithms. then, use a calculator to obtain a decimal approximation for the solution.
( e^{2x} - 14e^x + 13 = 0 )

the solution set expressed in terms of logarithms is (\boxed{quad})

use a comma to separate the answers as needed. simplify your answer. use integers or fractions for any numbers in the expression. use (ln) for natural logarithm and (log) for common logarithm.

Explanation:

Step1: Let \( y = e^x \), then the equation \( e^{2x}-14e^x + 13 = 0 \) becomes a quadratic equation \( y^2-14y + 13 = 0 \).

Step2: Factor the quadratic equation. We need two numbers that multiply to 13 and add up to -14. The numbers are -1 and -13. So, \( y^2-14y + 13=(y - 1)(y - 13)=0 \).

Step3: Solve for \( y \). Setting each factor equal to zero gives \( y - 1 = 0 \) or \( y - 13 = 0 \), so \( y = 1 \) or \( y = 13 \).

Step4: Substitute back \( y = e^x \). For \( y = 1 \), we have \( e^x=1 \), and since \( e^0 = 1 \), \( x = 0 \). For \( y = 13 \), we have \( e^x=13 \), and taking the natural logarithm of both sides gives \( x=\ln(13) \).

Answer:

The solution set is \( \{0, \ln(13)\} \) (or the decimal approximations are \( 0 \) and approximately \( 2.5649 \)).