Solve for X in the Fractional Power Equation: ((1/x)+(1/2))² / ((1/x)+(1/3))² = 81/64

$\frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}$

Find X

To solve this problem, we'll follow these steps:

• Step 1: Cross-multiply the given equation to eliminate fractions.
• Step 2: Expand the squared terms on either side of the equation.
• Step 3: Rearrange terms to form a quadratic equation.
• Step 4: Solve the quadratic equation to find possible values for $x$.

Now, let's work through each step:

Step 1: Begin with the given equation:
$\frac{(\frac{1}{x} + \frac{1}{2})^2}{(\frac{1}{x} + \frac{1}{3})^2} = \frac{81}{64}$. Cross-multiply to eliminate fractions:
$(\frac{1}{x} + \frac{1}{2})^2 \times 64 = (\frac{1}{x} + \frac{1}{3})^2 \times 81$.

Step 2: Expand each squared term:
For $(\frac{1}{x} + \frac{1}{2})^2$, use $(a + b)^2 = a^2 + 2ab + b^2$:
$(\frac{1}{x})^2 + 2(\frac{1}{x})(\frac{1}{2}) + (\frac{1}{2})^2 = \frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}$.
Similarly, $(\frac{1}{x} + \frac{1}{3})^2 = \frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}$.

Step 3: Substitute these into the cross-multiplied equation:
$64\left(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}\right) = 81\left(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}\right)$.

Step 4: Simplify and collect like terms:
$64(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}) = (64\frac{1}{x^2} + 64\frac{1}{x} + 16)$,
$81(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}) = (81\frac{1}{x^2} + 54\frac{1}{x} + 9)$.

Equating terms gives:
$64\frac{1}{x^2} + 64\frac{1}{x} + 16 = 81\frac{1}{x^2} + 54\frac{1}{x} + 9$.

Step 5: Solve the quadratic equation:
Combine like terms: $-17\frac{1}{x^2} + 10\frac{1}{x} + 7 = 0$.
Let $y = \frac{1}{x}$. Substitute to get: $-17y^2 + 10y + 7 = 0$.
Multiply the entire equation by -1 to simplify: $17y^2 - 10y - 7 = 0$.

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=17$, $b=-10$, $c=-7$:
$y = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 17 \times (-7)}}{2 \times 17}$
$y = \frac{10 \pm \sqrt{100 + 476}}{34}$
$y = \frac{10 \pm \sqrt{576}}{34}$
$y = \frac{10 \pm 24}{34}$
Which gives:
$y = \frac{34}{34} = 1$ or $y = -\frac{14}{34} = -\frac{7}{17}$.

Since $y = \frac{1}{x}$:
For $y=1$, $x = \frac{1}{y} = 1$.
For $y=-\frac{7}{17}$, $x = \frac{1}{y} = -\frac{17}{7}$.

Therefore, the solutions for $x$ are $x = 1$ and $x = -\frac{17}{7}$.

Checking the correct answer choice, these correspond to the second choice.

Thus, the solution to the problem is $x = 1, -\frac{17}{7}$.