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Q.
If the $6^{th}$ term in the expansion of $(\frac{1}{x^{8/3}}+x^{2} log _{10} x)^{8}$ is $5600$, then $x$ equals
Binomial Theorem
Solution:
It is given that $6^{th}$ term in the expansion of $(\frac{1}{x^{8/3}}+x^{2} log_{10}x)^{8}$ is $5600;$ therefore,
$^{8}C_{5}(x^{2} log_{10} x)^{5}(\frac{1}{x^{8/3}})^{3}=5600$
or $56x^{10} (log_{10} x)^{5} \frac{1}{x^{8}}=5600$
or $x^{2}(log_{10} x)^{5}=100$
or $x^{2} (log_{10} x)^{5}=10^{2} (log_{10}10)^{5}$
or $x=10$