Question

If $6^{th}$ term in the expansion of ${(\frac{1}{x^{8/3}}+x^2{\log }_{10}x)}^8$ is $5600$, then $x$ is equal to

• A. $8$
• B. $9$
• C. $10$
• D. $Noneofthese$

Answer 1

Answer: (C)

$10$

A] $(\frac{1}{x^{8/3}}+x^{2}lo{g}_{10}x{)}^8$.

$T_6=T_{5+1}={}^8{C}_5\left(\frac{1}{x^{8/3}}{)}^{8-5}\right(x^{2}lo{g}_{10}x{)}^5=5600$

$=56\left(\frac{1}{x^8}\right)x^{10}(lo{g}_{10}x{)}^5=5600$

$=x^2(lo{g}_{10}x{)}^5=100$

$=\therefore x=10$