We are tasked with solving the equation:

$\log_2(x + 1) = 2$

Step-by-Step Solution:

1. Rewrite the equation in exponential form: The equation $\log_b(y) = z$ can be rewritten as $y = b^z$, where $b$ is the base, $y$ is the argument of the logarithm, and $z$ is the exponent.

   Using this rule, we can rewrite the equation $\log_2(x + 1) = 2$ as:

   $x + 1 = 2^2$

2. Simplify the equation: Now, simplify the exponent:

   $x + 1 = 4$

3. Solve for $x$: Subtract 1 from both sides:

   $x = 4 - 1$ $x = 3$

Final Answer:

$x = 3$

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. What is the general formula for converting a logarithmic equation to an exponential equation?
2. How would the solution change if the base of the logarithm were different, say $\log_3(x + 1) = 2$?
3. How do we handle logarithmic equations when there are multiple logarithms in the same equation?
4. Can the logarithmic equation $\log_2(x + 1) = -2$ be solved? What would happen?
5. What are the properties of logarithms that can simplify solving complex equations?

Tip: When solving logarithmic equations, always check that the argument of the logarithm (the expression inside the log) is positive, as the logarithm of a negative number is undefined in real numbers.