The problem asks to find the value of $x$ in the equation $16 \times 2^{(x+1)} = 4^x \times 8^{(1-x)}$.

1. Problem Description

The problem asks to find the value of $x$ in the equation $16 \times 2^{(x+1)} = 4^x \times 8^{(1-x)}$.

2. Solution Steps

First, rewrite all the numbers as powers of 2:

$16 = 2^4$

$4 = 2^2$

$8 = 2^3$

So the equation becomes:

$2^4 \times 2^{(x+1)} = (2^2)^x \times (2^3)^{(1-x)}$

Using the property $a^m \times a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$, we have:

$2^{4+x+1} = 2^{2x} \times 2^{3(1-x)}$

$2^{x+5} = 2^{2x+3-3x}$

$2^{x+5} = 2^{3-x}$

Since the bases are equal, we can equate the exponents:

$x+5 = 3-x$

Add $x$ to both sides:

$2x+5 = 3$

Subtract 5 from both sides:

$2x = -2$

Divide by 2:

$x = -1$

3. Final Answer

D. -1