Let's solve the equation step by step:

Given:

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

Step 1: Combine logarithmic terms

We can use the property of logarithms to combine terms on the right-hand side. Notice that $\log 2$ is a common factor:

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

Step 2: Use the logarithmic property $\log a + \log b = \log(ab)$

Now, we can combine the right-hand side using the property of logarithms that says $\log a + \log b = \log(ab)$:

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

Since $2 \cdot 2^x = 2^{x+1}$, we have:

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

Step 3: Remove the logarithms

Now that we have $\log$ on both sides, we can drop the logarithms (because if $\log a = \log b$, then $a = b$):

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

Step 4: Solve the equation

At this point, the equation is:

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

This is not a straightforward linear equation, so we need to find a solution. Let's check for possible values of $x$ by trial and error.

Try $x = 1$:

$2(1) + 8 = 2^{1+1} \quad \Rightarrow \quad 2 + 8 = 4$

$10 \neq 4$

So $x = 1$ is not a solution.

Try $x = 2$:

$2(2) + 8 = 2^{2+1} \quad \Rightarrow \quad 4 + 8 = 8$

$12 = 8$

So $x = 2$ is not a solution either.

Try $x = 3$:

$2(3) + 8 = 2^{3+1} \quad \Rightarrow \quad 6 + 8 = 16$

$14 = 16$

This is still not a solution.

Would you like me to try a different method to solve for $x$? Let me know if you need more detailed steps!