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7.Binomial Theorem
normal
If the $6^{th}$ term in the expansion of the binomial ${\left[ {\frac{1}{{{x^{\frac{8}{3}}}}}\,\, + \,\,{x^2}\,{{\log }_{10}}\,x} \right]^8}$ is $5600$, then $x$ equals to
A
$5$
B
$8$
C
$10$
D
$100$
Solution
$T_6 = ^8C_5 {\left( {\frac{1}{{{x^{\frac{8}{3}}}}}} \right)^{8\, – \,5}}. (x^2 log_{10} x)^5 = 100 \Rightarrow x = 10$
Standard 11
Mathematics