Solve for x.
-$e ^ { 4 x ^ { 2 } }$ + $e ^ { ( 2 x ) ^ { 2 } }$ = 6
A) x = ± $\frac { \sqrt { \ln 3 } } { 2 }$
B) x = $\frac { 1 } { 2 }$ ln $\frac { 1 } { 2 }$
C) x = ± $\frac { 1 } { 4 }$ $\sqrt { \ln 6 }$
D) x = ± $\frac { 1 } { 2 }$ $\sqrt { \ln 6 }$

Question

Solve for x. 
-$e ^ { 4 x ^ { 2 } }$ + $e ^ { ( 2 x ) ^ { 2 } }$ = 6
A) x = ± $\frac { \sqrt { \ln 3 } } { 2 }$
B) x = $\frac { 1 } { 2 }$ ln $\frac { 1 } { 2 }$
C) x = ± $\frac { 1 } { 4 }$ $\sqrt { \ln 6 }$
D) x = ± $\frac { 1 } { 2 }$ $\sqrt { \ln 6 }$

Quizplus · Accepted Answer

The equation simplifies to $e^{4x^2} + e^{4x^2} = 6$ because $(2x)^2 = 4x^2$. Combining terms gives $2e^{4x^2} = 6$, so $e^{4x^2} = 3$. Taking the natural logarithm of both sides gives $4x^2 = \ln(3)$, and solving for x yields $x = ±\frac{\sqrt{\ln(3)}}{2}$.

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