Question

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

• A. 0
• B. $log_e 10 $
• C. 10
• D. x does not exist.

Answer 1

Answer: (C)

$T_{6} = T_{5+1} =\,{}^{8}C_{5} \left(\frac{1}{x^{8/3}}\right)^{2} \left(x^{2}\,log_{10}\,x\right)^{5} = 5600$
$\Rightarrow \frac{8 \times7\times 6}{1\times 2\times 3}. \frac{1}{x^{8}} x^{10} \left(log_{10}\,x\right)^{5}= 5600$
$\Rightarrow x^{2}\left(log_{10}\,x^{5} \right) = 100$
$\Rightarrow x^{2} \left(log_{10} x\right)^{5} = \left(10\right)^{2}\left(log_{10}\,10\right)^{5}$
$\Rightarrow x = 10$