Question

If the sixth term in the expansion of ${(\frac{1}{x^{8/3}}+x^2{\log }_{10}x)}^8$ is $5600$, find $x.$

Answer 1

$(\frac{1}{x^{\frac{8}{3}}}+x^2\log x{)}^8$

Its General term is

$T_{r+1}{=}^n{C}_r\left(\frac{1}{x^{\frac{8}{3}}}{)}^{8-r}\right(x^2\log x{)}^r$

So $6^{th}$ Term is $r+1=6$ $\Rightarrow r=5$

${}^8{C}_5\left(\frac{1}{x^{\frac{8}{3}}}{)}^3\right(x^2\log x{)}^5=5600$

$\Rightarrow x^{-8+10}{\left(\log x\right)}^5\times 8\times 7=5600$

$\Rightarrow x^2(\log x{)}^5=100$

$\Rightarrow x=10$