To solve the equation \( \ln(x) - \ln(x+2) = 1 \), follow these steps:

### Step 1: Use the logarithm subtraction property
The property of logarithms that states \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \) can be applied here:
\[
\ln(x) - \ln(x+2) = \ln\left(\frac{x}{x+2}\right)
\]
So, the equation becomes:
\[
\ln\left(\frac{x}{x+2}\right) = 1
\]

### Step 2: Exponentiate both sides
To eliminate the natural logarithm, exponentiate both sides with base \( e \) (the natural logarithm base):
\[
\frac{x}{x+2} = e^1
\]
\[
\frac{x}{x+2} = e
\]

### Step 3: Solve the equation for \( x \)
Multiply both sides by \( x + 2 \) to get rid of the fraction:
\[
x = e(x + 2)
\]
Expand the equation:
\[
x = ex + 2e
\]
Subtract \( ex \) from both sides:
\[
x - ex = 2e
\]
Factor out \( x \) on the left side:
\[
x(1 - e) = 2e
\]
Finally, solve for \( x \):
\[
x = \frac{2e}{1 - e}
\]

### Step 4: Simplify (optional)
Since the solution involves \( e \), you can leave it as:
\[
x = \frac{2e}{1 - e}
\]

### Answer:
The solution is \( x = \frac{2e}{1 - e} \).

Would you like more details on this solution or have any other questions?

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Here are 5 related questions you might find interesting:
1. How do you solve an equation involving logarithms with different bases?
2. What are the properties of logarithms, and how are they used in equations?
3. How does the change of base formula work in logarithms?
4. What is the relationship between logarithms and exponential functions?
5. How can you solve a logarithmic equation when the variable is inside the logarithm?

**Tip:** When working with logarithmic equations, always check that the argument of the logarithm remains positive, as logarithms of non-positive numbers are undefined in the real number system.

Learn how to solve the logarithmic equation ln(x) - ln(x+2) = 1 with a detailed step-by-step solution. This problem covers key concepts in logarithms and exponential equations, including the logarithm subtraction property and exponentiation. Suitable for advanced high school students and beyond.

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