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.

Learn how to solve the logarithmic equation log base 2 (x-1) + 1 = Log base 2 (x+3) step-by-step. This problem involves logarithms and equations, applying the logarithm subtraction rule to simplify and solve. Suitable for high school students in Grades 10-12.

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