We need to solve the equation:

$2 + \log \sqrt{1+x} + 3 \log \sqrt{1-x} = \log \sqrt{1-x^2}$

Step 1: Simplify the logarithmic terms.

Recall that $\log a^b = b \log a$, so we can simplify the logarithmic terms:

$\log \sqrt{1+x} = \frac{1}{2} \log(1+x)$ $3 \log \sqrt{1-x} = \frac{3}{2} \log(1-x)$

The equation becomes:

$2 + \frac{1}{2} \log(1+x) + \frac{3}{2} \log(1-x) = \frac{1}{2} \log(1-x^2)$

Step 2: Multiply the entire equation by 2 to eliminate the fractions.

$4 + \log(1+x) + 3 \log(1-x) = \log(1-x^2)$

Step 3: Use logarithmic properties to combine terms.

Use the property $\log a + \log b = \log(ab)$:

$4 + \log\left[(1+x)(1-x)^3\right] = \log(1-x^2)$

Step 4: Move the constant term.

Subtract 4 from both sides:

$\log\left[(1+x)(1-x)^3\right] = \log(1-x^2) - 4$

Step 5: Exponentiate both sides to eliminate the logarithm.

To remove the logarithms, we use the fact that if $\log a = \log b$, then $a = b$:

$(1+x)(1-x)^3 = \frac{1-x^2}{e^4}$

Step 6: Simplify the equation.

Expand $(1-x)^3$ and simplify the equation. But before we do that, we notice that $1-x^2 = (1-x)(1+x)$, which allows us to match the factors on both sides:

$(1+x)(1-x)^3 = \frac{(1-x)(1+x)}{e^4}$

Cancel $(1+x)$ from both sides:

$(1-x)^3 = \frac{1-x}{e^4}$

Further simplification leads to the possible solution for $x$ after canceling out the common terms:

$(1-x)^2 = \frac{1}{e^4}$

Taking the square root of both sides:

$1-x = \frac{1}{e^2}$

Solve for $x$:

$x = 1 - \frac{1}{e^2}$

Final Answer:

$x = 1 - \frac{1}{e^2}$

Would you like any details or have any questions?

---

Here are some related questions:

1. What is the base of natural logarithms?
2. How can logarithmic identities simplify complex equations?
3. What are the steps to solve logarithmic equations involving exponents?
4. How does one use properties of logarithms to combine or break down terms?
5. What are the applications of logarithmic equations in real-world problems?

Tip: When solving logarithmic equations, always remember to check for extraneous solutions by substituting back into the original equation.