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\
use a comma to separate 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 \( y = e^x \)

The equation \( e^{2x}-14e^x + 13 = 0 \) can be rewritten as a quadratic equation in terms of \( y \): \( y^2-14y + 13 = 0 \)

Step2: Solve the quadratic equation

Factor the quadratic equation: \( (y - 1)(y - 13)=0 \)
Set each factor equal to zero: \( y - 1 = 0 \) or \( y - 13 = 0 \)
So, \( y = 1 \) or \( y = 13 \)

Step3: Substitute back \( y = e^x \)

• For \( y = 1 \): \( e^x=1 \)

Take the natural logarithm of both sides: \( \ln(e^x)=\ln(1) \)
Since \( \ln(e^x)=x \) and \( \ln(1) = 0 \), we get \( x = 0 \)

• For \( y = 13 \): \( e^x=13 \)

Take the natural logarithm of both sides: \( \ln(e^x)=\ln(13) \)
Since \( \ln(e^x)=x \), we get \( x=\ln(13) \)

Answer:

The solution set in terms of logarithms is \( 0, \ln(13) \) (and the decimal approximations are \( 0, \approx 2.5649 \))