We are given the equation $\log_y(2x+3) + \log_y(2x-3) = 1$, and we are asked to find an expression for $y$ in terms of $x$.

1. Problem Description

We are given the equation $\log_y(2x+3) + \log_y(2x-3) = 1$, and we are asked to find an expression for $y$ in terms of $x$.

2. Solution Steps

We can use the logarithm product rule to combine the two logarithm terms:

$\log_a(b) + \log_a(c) = \log_a(bc)$

Applying this rule to the given equation, we get:

$\log_y((2x+3)(2x-3)) = 1$

The expression $(2x+3)(2x-3)$ is in the form $(a+b)(a-b)$, which simplifies to $a^2 - b^2$. Therefore,

$(2x+3)(2x-3) = (2x)^2 - (3)^2 = 4x^2 - 9$

So the equation becomes:

$\log_y(4x^2 - 9) = 1$

Now, we can rewrite the logarithmic equation in exponential form. The general rule is:

$\log_a(b) = c \Leftrightarrow a^c = b$

Applying this rule to our equation, we get:

$y^1 = 4x^2 - 9$

Therefore,

$y = 4x^2 - 9$

3. Final Answer

$y = 4x^2 - 9$