We are given the equation:

$2 \log_2 \left( \sqrt{x} \right) + \log_2 (x + 3) = 2$

Step 1: Simplify the first logarithmic term

Using the property of logarithms $\log_b \left( a^n \right) = n \log_b(a)$, we can simplify the first term:

$2 \log_2 \left( \sqrt{x} \right) = 2 \log_2 \left( x^{1/2} \right) = \log_2 (x)$

Thus, the equation becomes:

$\log_2 (x) + \log_2 (x + 3) = 2$

Step 2: Use the logarithmic addition property

We apply the property $\log_b(a) + \log_b(c) = \log_b(ac)$ to combine the logarithms:

$\log_2 \left( x(x + 3) \right) = 2$

Step 3: Convert from logarithmic form to exponential form

Recall that $\log_b(a) = c$ implies $a = b^c$. Using this here:

$x(x + 3) = 2^2$

$x(x + 3) = 4$

Step 4: Expand and solve the quadratic equation

Now, expand and simplify:

$x^2 + 3x = 4$

$x^2 + 3x - 4 = 0$

Step 5: Factor the quadratic equation

We now factor the quadratic:

$(x + 4)(x - 1) = 0$

Step 6: Solve for $x$

Set each factor equal to zero:

$x + 4 = 0 \quad \text{or} \quad x - 1 = 0$

Thus,

$x = -4 \quad \text{or} \quad x = 1$

Step 7: Check for valid solutions

We need to check if these values are valid in the original logarithmic equation. Logarithms are only defined for positive arguments, so we must have $x > 0$ and $x + 3 > 0$.

• For $x = -4$, $\log_2 \left( \sqrt{-4} \right)$ is undefined because the square root of a negative number is not real.
• For $x = 1$, all terms are valid, so $x = 1$ is the solution.

Final Answer:

The solution is $x = 1$.

---

Would you like to see a more detailed breakdown of any of the steps or have other questions?

Here are some related questions to explore further:

1. How do you handle logarithmic equations with more complex exponents?
2. What are the other properties of logarithms that can simplify equations?
3. How do you solve logarithmic equations with different bases?
4. What happens if the argument of a logarithm is zero or negative?
5. How do you apply the change of base formula in logarithmic equations?

Tip: Always remember to check the domain of your solution when working with logarithms!