Question

If the 6th term in the expansion of${\left(\frac{1}{x^{8/3}}+x^{2}lo{g}_{10}x\right)}^8$ is 5600, then x equals

• A. 0
• B. $lo{g}_{e}10$
• C. 10
• D. x does not exist.

Answer 1

Answer: (C)

$T_6=T_{5+1}={}^8{C}_5{\left(\frac{1}{x^{8/3}}\right)}^2{\left(x^{2}lo{g}_{10}x\right)}^5=5600$
$\Rightarrow \frac{8\times 7\times 6}{1\times 2\times 3}.\frac{1}{x^8}{x}^{10}{\left(lo{g}_{10}x\right)}^5=5600$
$\Rightarrow x^2\left(lo{g}_{10}{x}^5\right)=100$
$\Rightarrow x^2{\left(lo{g}_{10}x\right)}^5={\left(10\right)}^2{\left(lo{g}_{10}10\right)}^5$
$\Rightarrow x=10$