To solve the inequality:

(√(x+3)/log2 (2x-3)) <= 0

First, we need to find the critical points where the function is equal to zero or undefined.

The numerator, √(x+3), will be equal to zero when x = -3.

The denominator, log2(2x-3), will be equal to zero when 2x-3 = 1, or x = 2.

Now, we need to test the intervals between the critical points -∞, -3, 2, and +∞.

Test the interval (-∞, -3):

Choose x = -4:
(√(-4+3)/log2(2(-4)-3)) = (√(-1)/log2(-11))

Since both the numerator and denominator are not negative in this interval, the function is greater than zero in this interval.

Test the interval (-3, 2):

Choose x = 0:
(√(0+3)/log2(2(0)-3)) = (√3/log2(-3))

Since the numerator is positive and the denominator is negative, the function is less than zero in this interval.

Test the interval (2, +∞):

Choose x = 3:
(√(3+3)/log2(2(3)-3)) = (√6/log2(3))

Both the numerator and the denominator are positive in this interval, so the function is greater than zero.

Therefore, the solution to the inequality is:

x ∈ (-3, 2]