To solve the equation:
$\log_2 x + \log_2 (x+7) = 3$

Step 1: Use the Logarithm Rule

We use the log addition rule:
$\log_b A + \log_b B = \log_b (A \times B)$
Applying this rule:
$\log_2 (x(x+7)) = 3$
which simplifies to:
$\log_2 (x^2 + 7x) = 3$

Step 2: Convert to Exponential Form

By definition, $\log_b A = C$ means:
$A = b^C$
So:
$x^2 + 7x = 2^3$
$x^2 + 7x = 8$

Step 3: Solve the Quadratic Equation

Rearrange:
$x^2 + 7x - 8 = 0$
Factorize:
$(x + 8)(x - 1) = 0$
Solve for $x$:
$x + 8 = 0 \quad \Rightarrow \quad x = -8$
$x - 1 = 0 \quad \Rightarrow \quad x = 1$

Step 4: Check for Validity

Logarithms are undefined for negative or zero inputs.
Since $\log_2(-8)$ is not valid, we discard $x = -8$.
Thus, the only solution is:
$\boxed{1}$

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Do you need more explanation or have any questions?

Here are some related problems to try:

1. Solve $\log_3 x + \log_3 (x+2) = 2$.
2. Solve $\log_5 (x-1) + \log_5 (x+3) = 1$.
3. Solve $\log_4 (x^2 - 3x) = 2$.
4. Solve $\log_2 (x+1) - \log_2 (x-1) = 1$.
5. Solve $2\log_2 x = 4$.

Tip:

Always check your solutions to logarithmic equations to avoid invalid values (negative or zero inside the log).