The problem is to solve the equation $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$ for $x$.

1. Problem Description

The problem is to solve the equation $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$ for $x$.

2. Solution Steps

The given equation is $2^{3x+1} - 3(2^{2x}) + 2^{x+1} = 2^x$. We can rewrite the terms using the properties of exponents:

$2^{3x+1} = 2^{3x} \cdot 2^1 = 2 \cdot (2^x)^3$

$2^{2x} = (2^x)^2$

$2^{x+1} = 2^x \cdot 2^1 = 2 \cdot 2^x$

Substituting these into the equation, we have:

$2(2^x)^3 - 3(2^x)^2 + 2(2^x) = 2^x$

Let $y = 2^x$. Then the equation becomes:

$2y^3 - 3y^2 + 2y = y$

$2y^3 - 3y^2 + y = 0$

$y(2y^2 - 3y + 1) = 0$

$y(2y - 1)(y - 1) = 0$

So the possible values for $y$ are $y = 0$, $y = \frac{1}{2}$, and $y = 1$.

Since $y = 2^x$, we have:

$2^x = 0$, $2^x = \frac{1}{2}$, and $2^x = 1$.

$2^x = 0$ has no solution, because $2^x$ is always positive.

$2^x = \frac{1}{2} = 2^{-1}$ implies $x = -1$.

$2^x = 1 = 2^0$ implies $x = 0$.

Therefore, the solutions for $x$ are $x = -1$ and $x = 0$.

3. Final Answer

$x = -1, 0$