The solution set of the equation $8 ^ { 2 x - 3 } = \left( \frac { 1 } { 32 } \right) ^ { x + 5 }$ is

A) $\left\{ - \frac { 16 } { 11 } \right\}$
B) $\left\{ \frac { 8 } { 11 } \right\}$
C) $\left\{ \frac { 4 } { 11 } \right\}$
D) $\left\{ - \frac { 2 } { 11 } \right\}$
E) $\varnothing$

Question

The solution set of the equation $8 ^ { 2 x - 3 } = \left( \frac { 1 } { 32 } \right) ^ { x + 5 }$ is

A) $\left\{ - \frac { 16 } { 11 } \right\}$
B) $\left\{ \frac { 8 } { 11 } \right\}$
C) $\left\{ \frac { 4 } { 11 } \right\}$
D) $\left\{ - \frac { 2 } { 11 } \right\}$
E) $\varnothing$

Quizplus · Accepted Answer

To solve the equation $8 ^ { 2 x - 3 } = \left( \frac { 1 } { 32 } \right) ^ { x + 5 }$, we can express both sides of the equation with a common base. The base can be 2, as $8 = 2^3$ and $32 = 2^5$. Thus, the equation becomes $2^{3(2x-3)} = \left(2^{-5}\right)^{x+5}$. Simplifying both sides gives $2^{6x-9} = 2^{-5x-25}$. Since the bases are the same, we can set the exponents equal to each other: $6x - 9 = -5x - 25$. Solving for $x$ gives $x = -\frac{16}{11}$.

The Solution Set of the Equation 82x−3=(132)x+58 ^ { 2 x - 3 } = \left( \frac { 1 } { 32 } \right) ^ { x + 5 }82x−3=(321​)x+5

The Solution Set of the Equation $8 ^ { 2 x - 3 } = \left( \frac { 1 } { 32 } \right) ^ { x + 5 }$