Let's solve the equation step by step:

Given:

$\log(2x + 8) = \log 2 + x \log 2$

Step 1: Simplify the Right-Hand Side

The right-hand side has two terms, $\log 2$ and $x \log 2$. You can factor out $\log 2$:

$\log(2x + 8) = \log 2 (1 + x)$

Step 2: Use the property of logarithms

Since both sides of the equation involve logarithms, we can assume the bases are the same (i.e., base 10) and apply the property $\log a = \log b \implies a = b$ (if the logs are of the same base).

Thus, we can equate the arguments inside the logarithms:

$2x + 8 = 2^{(1 + x)}$

Step 3: Solve the equation

Now, we need to solve the equation:

$2x + 8 = 2^{(x + 1)}$

First, subtract 8 from both sides:

$2x = 2^{(x + 1)} - 8$

Step 4: Trial solutions (since this is not easily solvable algebraically)

At this point, it might be easiest to check a few values of $x$ to find a solution.

Let’s check some small values of $x$:

• For $x = 0$:

$2(0) + 8 = 8 \quad \text{and} \quad 2^{(0 + 1)} = 2^1 = 2$ So, $8 \neq 2$, so $x = 0$ is not a solution.

• For $x = 2$:

$2(2) + 8 = 4 + 8 = 12 \quad \text{and} \quad 2^{(2 + 1)} = 2^3 = 8$ Again, $12 \neq 8$, so $x = 2$ is not a solution.

• For $x = 3$:

$2(3) + 8 = 6 + 8 = 14 \quad \text{and} \quad 2^{(3 + 1)} = 2^4 = 16$ So, $14 \neq 16$, so $x = 3$ is not a solution either.

• For $x = 4$:

$2(4) + 8 = 8 + 8 = 16 \quad \text{and} \quad 2^{(4 + 1)} = 2^5 = 32$ Again, the equation does not hold.

Let me know if you would like to proceed differently.