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 \( y \):
- \( y = \frac{16}{2} = 8 \)
- \( y = \frac{-8}{2} = -4 \) (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:
\[ \boxed{1} \]
