Sketch the graph of of $y = \log_2(3x + 1)$ and label the point where $x = \frac{1}{3}$. Explain why $x > -\frac{1}{3}$. Solve $\log_2(3x + 1) + \frac{1}{2}\log_{\sqrt{2}}(3x - 1) = 1$.

1. Identify the vertical asymptote and domain of the logarithmic function

For $y = \log_2(3x + 1)$, the argument must be positive: $3x + 1 > 0 \implies x > -\frac{1}{3}$.
The vertical asymptote is $x = -\frac{1}{3}$.

2. Find the coordinates of the required point and the axis intercepts

When $x = \frac{1}{3}$, $y = \log_2(3(\frac{1}{3}) + 1) = \log_2(2) = 1$. The point is $(\frac{1}{3}, 1)$.
When $y = 0$, $\log_2(3x + 1) = 0 \implies 3x + 1 = 1 \implies x = 0$. The $x$-intercept is $(0, 0)$.

3. Sketch the graph and explain the domain restriction

Sketch a curve passing through $(0,0)$ and $(\frac{1}{3}, 1)$, approaching the vertical line $x = -\frac{1}{3}$ as $x \to -\frac{1}{3}^+$, and increasing logarithmically for $x > 0$.
Explanation: The argument of a logarithm must be strictly positive. Therefore, $3x + 1 > 0$, which gives $x > -\frac{1}{3}$.

1. Change the base of the second logarithmic term to 2.

$\log_{\sqrt{2}}(3x - 1) = \frac{\log_2(3x - 1)}{\log_2(\sqrt{2})} = \frac{\log_2(3x - 1)}{\frac{1}{2}} = 2\log_2(3x - 1)$

2. Substitute back into the equation and simplify.

$\log_2(3x + 1) + \frac{1}{2}[2\log_2(3x - 1)] = 1$
$\log_2(3x + 1) + \log_2(3x - 1) = 1$

3. Combine the logarithms using the product law.

$\log_2[(3x + 1)(3x - 1)] = 1$
$\log_2(9x^2 - 1) = 1$

4. Convert to exponential form and solve for $x$.

$9x^2 - 1 = 2^1$
$9x^2 = 3$
$x^2 = \frac{1}{3}$
$x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3}$

5. Check the domain of the original logarithms to reject invalid solutions.

For $\log_{\sqrt{2}}(3x - 1)$ to be defined, $3x - 1 > 0 \implies x > \frac{1}{3}$.
Since $-\frac{\sqrt{3}}{3} < \frac{1}{3}$, we reject $x = -\frac{\sqrt{3}}{3}$.
Therefore, $x = \frac{\sqrt{3}}{3}$.