Solve the equation.
-$$e ^ { 2 x } - 14 e ^ { x } + 9 = 0$$
A) { }
B) $\{ \ln ( 7 \pm 2 \sqrt { 10 } ) \} ; x \approx 2.5896 , x \approx - 0.3924$
C) $\{ \ln ( 7 + 2 \sqrt { 10 } ) \} ; x \approx 2.5896$
D) $\{ 7 \pm 2 \sqrt { 10 } \} ; x \approx 13.3246 , x \approx 0.6754$

Question

Solve the equation.
-$$e ^ { 2 x } - 14 e ^ { x } + 9 = 0$$
A) {  }
B) $\{ \ln ( 7 \pm 2 \sqrt { 10 } ) \} ; x \approx 2.5896 , x \approx - 0.3924$
C) $\{ \ln ( 7 + 2 \sqrt { 10 } ) \} ; x \approx 2.5896$
D) $\{ 7 \pm 2 \sqrt { 10 } \} ; x \approx 13.3246 , x \approx 0.6754$

Quizplus · Accepted Answer

This equation is a quadratic in form if we let $y = e^x$, transforming it into $y^2 - 14y + 9 = 0$. Solving this quadratic equation gives $y = 7 \pm 2\sqrt{10}$. Substituting back $e^x$ for $y$ and solving for $x$ gives $x = \ln(7 \pm 2\sqrt{10})$, which corresponds to the solutions provided in option B.

Solve the Equation e2x−14ex+9=0e ^ { 2 x } - 14 e ^ { x } + 9 = 0e2x−14ex+9=0

Solve the Equation $e ^ { 2 x } - 14 e ^ { x } + 9 = 0$