3^y+1 /32 = root96^x

To solve for y in terms of x, we can rewrite the equation as follows:

\(3^{y+1} / 32 = \sqrt{96^x}\)

First, simplify the right side of the equation:

\(\sqrt{96^x} = \sqrt{(3^2 \cdot 2)^x} = \sqrt{3^{2x} \cdot 2^x} = 3^x \cdot \sqrt{2^x} = 3^x \cdot 2^{x/2}\)

Now, substitute this back into the original equation:

\(3^{y+1} / 32 = 3^x \cdot 2^{x/2}\)

Next, we can rewrite 32 as \(2^5\):

\(3^{y+1} / 2^5 = 3^x \cdot 2^{x/2}\)

Now, we can simplify the equation by dividing both sides by \(3^x\) and \(2^5\):

\(3^{y-x+1} = 2^{(3x/2) - 5}\)

Since the bases are equal, we can set the exponents equal to each other:

\(y-x+1 = (3x/2) - 5\)

Now, we can solve for y in terms of x:

\(y = (3x/2) - 5 + x - 1 = (5x/2) - 6\)

Therefore, the solution for y in terms of x is:

\[ y = \frac{5x}{2} - 6 \]