\( ( 2 ^ { x + 1 } ) ^ { 4 } \cdot 2 ^ { 1 - \partial x } = 1728 \)

To solve the equation \( (2^{x + 1})^4 \cdot 2^{1 - \partial x} = 1728 \), we can simplify it step by step. First, we combine the exponents on the left side: \[ (2^{x + 1})^4 = 2^{4(x + 1)} = 2^{4x + 4} \] Thus, the equation becomes: \[ 2^{4x + 4} \cdot 2^{1 - \partial x} = 1728 \] Next, use the property of exponents that states \( a^m \cdot a^n = a^{m+n} \): \[ 2^{(4x + 4) + (1 - \partial x)} = 1728 \] This simplifies to: \[ 2^{4x + 5 - \partial x} = 1728 \] Now, convert 1728 into a power of 2. We calculate: \[ 1728 = 12^3 = (2^2 \cdot 3)^3 = 2^6 \cdot 3^3 \] Thus, \( 1728 \) is not purely a power of two, so let’s break it down in terms of the base 2. It can also be expressed as \( 2^6 \cdot 3^3 \). We need the left side to equal \( 2^{4x + 5 - \partial x} \) on its own. To find the components where \( 2^k \) must equal \( 2^6 \): \[ 4x + 5 - \partial x = 6 \] Now, rearranging gives us: \[ 4x - \partial x = 1 \] For \( \partial x \), it typically symbolizes differentiation, but here we treat it as a constant to be determined in the context of solving for \( x \). If we assume \( \partial x = n \) for our consideration: Let’s express \( n = k \). Then, the equation: \[ 4x - k = 1 \implies 4x = k + 1 \implies x = \frac{k + 1}{4} \] Thus, with \( k \) clarifying how \( \partial x \) behaves, we can substitute back to determine a specified real solution. Each solution will lead to a valid \( x \) based on any \( k \) used. To summarize, 1. Set \( 4x - k = 1, \) solve \( x \) in terms of \( k \). 2. Seek integer values for \( k \) applicable to your context or specific problem. This provides a route to finding \( x \) based on your variables.