Solve the equation.
-$$3 ^ { 2 x } + 3 ^ { x } - 6 = 0$$
A) $\{ \ln 6 \}$
B) $\left\{ \frac { \ln 2 } { \ln 3 } \right\}$
C) $\left\{ \frac { \ln 2 } { \ln 6 } \right\}$
D) $\left\{ \frac { \ln 3 } { \ln 2 } \right\}$

Question

Quizplus · Accepted Answer

To solve the equation $3^{2x} + 3^x - 6 = 0$, let $y = 3^x$. This transforms the equation into a quadratic equation: $y^2 + y - 6 = 0$. Factoring this gives $(y + 3)(y - 2) = 0$, so $y = -3$ or $y = 2$. Since $y = 3^x$ and must be positive, we discard $y = -3$ and keep $y = 2$. Thus, $3^x = 2$, and taking the natural logarithm of both sides gives $x \ln 3 = \ln 2$, or $x = \frac{\ln 2}{\ln 3}$.

Solve the Equation 32x+3x−6=03 ^ { 2 x } + 3 ^ { x } - 6 = 032x+3x−6=0

Solve the Equation $3 ^ { 2 x } + 3 ^ { x } - 6 = 0$