8^(2/х)-2^((2х+3)/х)-32=0

To solve the equation \( 8^{\frac{2}{x}} - 2^{\frac{2x + 3}{x}} - 32 = 0 \), we can start by expressing all terms with the same base. We know that \( 8 = 2^3 \), so we can rewrite the first term: \[ 8^{\frac{2}{x}} = (2^3)^{\frac{2}{x}} = 2^{\frac{6}{x}} \] Now let's look at the second term \( 2^{\frac{2x + 3}{x}} \): \[ 2^{\frac{2x + 3}{x}} = 2^{2 + \frac{3}{x}} \] Substituting back into the original equation gives: \[ 2^{\frac{6}{x}} - 2^{2 + \frac{3}{x}} - 32 = 0 \] Next, we can express \( 32 \) as a power of \( 2 \): \[ 32 = 2^5 \] Now we rewrite the equation: \[ 2^{\frac{6}{x}} - 2^{2 + \frac{3}{x}} - 2^5 = 0 \] To simplify the equation, let's set \( y = 2^{\frac{3}{x}} \). Then we have: \[ 2^{\frac{6}{x}} = (2^{\frac{3}{x}})^2 = y^2 \] and \[ 2^{2 + \frac{3}{x}} = 2^2 \cdot 2^{\frac{3}{x}} = 4y \] Now substituting these back into the equation results in: \[ y^2 - 4y - 32 = 0 \] This is a quadratic equation and can be solved using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -4, c = -32 \). Calculating the discriminant: \[ b^2 - 4ac = (-4)^2 - 4(1)(-32) = 16 + 128 = 144 \] Now we can find \( y \): \[ y = \frac{4 \pm \sqrt{144}}{2} \] \[ y = \frac{4 \pm 12}{2} \] This gives us two potential solutions for \( y \): 1. \( y = \frac{16}{2} = 8 \) 2. \( y = \frac{-8}{2} = -4 \) (not possible since \( y \) must be positive) Since \( y = 2^{\frac{3}{x}} \), set \( y = 8 \): \[ 2^{\frac{3}{x}} = 8 \implies 2^{\frac{3}{x}} = 2^3 \] Thus, we have: \[ \frac{3}{x} = 3 \implies x = 1 \] Therefore, the solution to the equation is: \[ \boxed{1} \]