Explanation

(d)
Note that for $\log_{10}x$ to be defined, $x > 0$
[image:8_image1.emf]
We have, $T_{6} = T_{5 + 1} = \ ^{8}C_{5}\left( \frac{1}{x^{\frac{8}{3}}} \right)^{8 - 5}\left( x^{2}\log_{10}x \right)^{5}$
$\Rightarrow \ 5600 = \frac{8!}{5!3!}\left( \frac{1}{x^{8}} \right)x^{10}{(\log_{10}x)}^{5}$
$\Rightarrow \ 5600 = 56x^{2}{(\log_{10}x)}^{5}$
$\Rightarrow \ 100 = x^{2}\left( \log_{10}x \right)^{5}$
$\Rightarrow \ \frac{100}{x^{2}} = {(\log_{10}x)}^{5}$
Let $y = \frac{100}{x^{2}}$
$\therefore\ y = \left( \log_{10}x \right)^{5}$
From the figure it is clear that curves intersect in just one point.
This point is $(10,\ 1)$
Therefore, $x = 10$