We are solving the equation:

\[
\log_2(x + 5) = 2 + \log_2(x + 2)
\]

### Step 1: Simplify using logarithmic properties
We know the logarithmic property:
\[
\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)
\]
Rewriting the equation, we isolate the logarithmic terms:
\[
\log_2(x + 5) - \log_2(x + 2) = 2
\]

Now apply the property:
\[
\log_2\left(\frac{x + 5}{x + 2}\right) = 2
\]

### Step 2: Eliminate the logarithm
To eliminate the logarithm, rewrite the equation in exponential form:
\[
\frac{x + 5}{x + 2} = 2^2
\]
\[
\frac{x + 5}{x + 2} = 4
\]

### Step 3: Solve for \(x\)
Multiply both sides by \(x + 2\) (ensuring \(x + 2 \neq 0\)):
\[
x + 5 = 4(x + 2)
\]
Expand the right-hand side:
\[
x + 5 = 4x + 8
\]
Simplify:
\[
5 - 8 = 4x - x
\]
\[
-3 = 3x
\]
\[
x = -1
\]

### Step 4: Verify the solution
Substitute \(x = -1\) into the original equation to ensure it satisfies the domain of the logarithms:
- For \(\log_2(x + 5)\), we have \(\log_2(-1 + 5) = \log_2(4)\), which is valid.
- For \(\log_2(x + 2)\), we have \(\log_2(-1 + 2) = \log_2(1)\), which is valid.

Now substitute into the original equation:
\[
\log_2(-1 + 5) = 2 + \log_2(-1 + 2)
\]
\[
\log_2(4) = 2 + \log_2(1)
\]
\[
2 = 2
\]

The solution is verified.

### Final Answer:
\[
x = -1
\]

Do you have any questions or want a detailed explanation of any step?

---

1. What are logarithmic properties, and how are they applied?
2. How do you handle domains when solving logarithmic equations?
3. Why is verifying solutions important in logarithmic equations?
4. What happens if a solution is not in the domain of the logarithm?
5. Can you explain how to rewrite logarithmic equations in exponential form?

**Tip:** Always check the domain of logarithmic functions to ensure solutions are valid.

Learn how to solve the logarithmic equation log2(x+5) = 2 + log2(x+2) step by step. This problem involves properties of logarithms, rewriting logarithmic expressions in exponential form, and solving for x. Ideal for Grades 10-12 students seeking to master logarithmic equations.

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