Question

If for a particular value of the variable \(x\), the following holds true: \(17 = \dfrac{17x}{1-x}\), then find the value of \(x^{2x}\).

• A. 17
• B. 1
• C. 2
• D. 1/2

Hint:

Simplify the equation first, the 17s cancel, solve for x, then substitute into the exponent expression.

Answer 1

The Correct Option is D

We can also solve this by first confirming the value of x satisfies the original equation with actual numbers, and then computing the required power step by step.

1. Start with $17 = \dfrac{17x}{1-x}$. Cross-multiply directly: $17(1-x) = 17x$.
2. Expand the left side: $17 - 17x = 17x$.
3. Bring like terms together: $17 = 17x + 17x = 34x$.
4. Solve for x: $x = \dfrac{17}{34} = \dfrac{1}{2}$.
5. Check this value in the original equation: left side = 17. Right side = $\dfrac{17 \times \frac{1}{2}}{1 - \frac{1}{2}} = \dfrac{8.5}{0.5} = 17$. Both sides match, confirming $x = \dfrac{1}{2}$ is correct.
6. Now compute $x^{2x}$. First find the exponent: $2x = 2 \times \dfrac{1}{2} = 1$. Then raise the base to this exponent: $x^{2x} = \left(\dfrac{1}{2}\right)^{1} = \dfrac{1}{2}$.

The value of $x^{2x}$ is $\dfrac{1}{2}$, matching option (4). \[\boxed{\dfrac{1}{2}}\]