The given equation is:

$\frac{\ln(x)^2}{3} = y$

To solve for $x$, we can follow these steps:

Step 1: Multiply both sides by 3

$\ln(x)^2 = 3y$

Step 2: Take the square root of both sides

$\ln(x) = \pm \sqrt{3y}$

Step 3: Solve for $x$ by exponentiating both sides using the base $e$

$x = e^{\pm \sqrt{3y}}$

Thus, the solution for $x$ is: $x = e^{\sqrt{3y}} \quad \text{or} \quad x = e^{-\sqrt{3y}}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How does the solution change if the logarithmic base is not $e$ but another number?
2. Can you explain why we take both positive and negative roots when solving?
3. What are the properties of logarithmic and exponential functions used in this solution?
4. How would you solve the equation if it involved a logarithm to a different base?
5. What happens to the solution if $y$ is negative?

Tip: When working with logarithmic equations, always check for the domain of the logarithmic function (in this case, $x > 0$).