The equation in the image is:

$\log_3{x} + \log_3{(x+2)} = 1$

Step-by-step solution:

1. Apply the logarithm property: The sum of two logarithms with the same base can be rewritten as the logarithm of the product:

   $\log_3(x) + \log_3(x+2) = \log_3[x(x+2)]$

   This simplifies to:

   $\log_3[x(x+2)] = 1$

2. Exponentiate both sides: Since $\log_3$ is the logarithm base 3, we can exponentiate both sides to remove the logarithm:

   $3^{\log_3[x(x+2)]} = 3^1$

   This simplifies to:

   $x(x + 2) = 3$

3. Expand and simplify: Expand the left side:

   $x^2 + 2x = 3$

   Subtract 3 from both sides to set the equation equal to 0:

   $x^2 + 2x - 3 = 0$

4. Factor the quadratic: Factor the quadratic equation:

   $(x + 3)(x - 1) = 0$

5. Solve for $x$: Set each factor equal to 0:

   $x + 3 = 0 \quad \text{or} \quad x - 1 = 0$

   Solving these gives:

   $x = -3 \quad \text{or} \quad x = 1$

6. Check for extraneous solutions: Logarithms are only defined for positive arguments, so we must check if these values of $x$ are valid in the original equation.

   • For $x = 1$, both $\log_3{1}$ and $\log_3{(1 + 2)}$ are defined.
   • For $x = -3$, $\log_3{(-3)}$ is not defined because logarithms of negative numbers are undefined in the real number system.

   Therefore, the only valid solution is $x = 1$.

Final Answer:

The solution is $x = 1$, which matches the selected answer.

Do you have any questions or would you like further details on any step?

Here are 5 related questions for practice:

1. How would the equation change if the logarithm base was 10 instead of 3?
2. Solve the equation $\log_5{x} + \log_5{(x+4)} = 1$.
3. Can you solve $\log_2{x} + \log_2{(x+3)} = 2$?
4. What happens if we have $\log_3{(x-2)} + \log_3{(x+5)} = 2$?
5. What are the properties of logarithms that help simplify log expressions?

Tip: Always check if your logarithmic solutions make the arguments inside the log functions positive to avoid extraneous solutions.