Let's solve the equation:

$\log_2{(x-1)} + 1 = \log_2{(x+3)}$

Step 1: Isolate the logarithmic terms.

First, subtract 1 from both sides:

$\log_2{(x-1)} = \log_2{(x+3)} - 1$

Step 2: Express the 1 as a logarithm.

We know that $1 = \log_2{2}$, so:

$\log_2{(x-1)} = \log_2{(x+3)} - \log_2{2}$

Step 3: Use the logarithm subtraction rule.

The subtraction of logarithms can be expressed as the logarithm of a division:

$\log_2{(x-1)} = \log_2{\left(\frac{x+3}{2}\right)}$

Step 4: Set the arguments equal to each other.

Since the logarithms are equal, their arguments must be equal:

$x - 1 = \frac{x + 3}{2}$

Step 5: Solve the equation.

Multiply both sides by 2 to eliminate the fraction:

$2(x - 1) = x + 3$

Expand and simplify:

$2x - 2 = x + 3$

Subtract $x$ from both sides:

$x - 2 = 3$

Add 2 to both sides:

$x = 5$

Step 6: Check the solution.

Substitute $x = 5$ back into the original equation:

$\log_2{(5-1)} + 1 = \log_2{(5+3)}$

$\log_2{4} + 1 = \log_2{8}$

$\log_2{4} = 2$ and $\log_2{8} = 3$:

$2 + 1 = 3$

The equation holds true, so the solution is:

$\boxed{5}$

Would you like more details on any step, or do you have any questions?

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Here are five related questions:

1. How do you solve logarithmic equations with different bases?
2. What are the properties of logarithms that are essential for solving equations?
3. Can you solve $\log_2{(x^2-1)} = \log_2{(x+2)}$?
4. What happens if there is no solution to a logarithmic equation?
5. How do you solve an equation where logarithms appear on both sides with different coefficients?

Tip: Always check your solutions by substituting them back into the original equation, especially with logarithmic equations, to ensure they don't result in taking the logarithm of a non-positive number.