If the 0th term in the expansion of $\left(\frac{1}{x} + x^2\log x\right)^8$ is 6000, then the value of $x$ is
Step-by-Step Solution
Key Concept: Exponential equations involving logarithms are solved by converting to equivalent algebraic forms and isolating the variable.
Setting $T_k = 56(\frac{1}{2})(2^{\log_{10}x})^2 = 56 \cdot 2^{(\log_{10}x)^2}$ and $T_k = 5600$ gives $50 \cdot 2^{(\log_{10}x)^2} = 5600$, so $2^{(\log_{10}x)^2} = 100$. Taking logarithms: $x^2(\log_{10}x)^2 = 100$, which means $x(\log_{10}x) = 10$, giving $x = 10$.
Correct Answer: 10