Solve the exponential equation below algebraically. Round your result to three decimal places.
$e ^ { 2 x } + 3 e ^ { x } - 18 = 0$
A) $1.099$
B) $0.214$
C) $2.077$
D) $- 0.188$
E) $3.271$

Question

Solve the exponential equation below algebraically. Round your result to three decimal places. 
$e ^ { 2 x } + 3 e ^ { x } - 18 = 0$
A) $1.099$
B) $0.214$
C) $2.077$
D) $- 0.188$
E) $3.271$

Quizplus · Accepted Answer

To solve the equation $e^{2x} + 3e^x - 18 = 0$, let $u = e^x$. Then the equation becomes $u^2 + 3u - 18 = 0$. Solving this quadratic equation using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find $u = 3$ or $u = -6$. Since $u = e^x$ must be positive, $u = 3$. Thus, $e^x = 3$, and solving for $x$ gives $x = \ln(3) \approx 1.099$.

Solve the Exponential Equation Below Algebraically e2x+3ex−18=0e ^ { 2 x } + 3 e ^ { x } - 18 = 0e2x+3ex−18=0

Solve the Exponential Equation Below Algebraically $e ^ { 2 x } + 3 e ^ { x } - 18 = 0$