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. $5$
• B. $4$
• C. $8$
• D. $noneofthese$

Answer 1

Answer: (D)

$noneofthese$

Using binomial theorem, the 6th term of $(a+b{)}^8$ is $56a^3{b}^5$

Now substituting the given terms, we get

$56\cdot \frac{1}{(x^{8/3}{)}^3}\cdot x^{10}\cdot (\log x{)}^5=5600$

$56\cdot \frac{1}{x^8}\cdot x^{10}\cdot (\log x{)}^5=5600$

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

$x^2(\log x{)}^5={10}^2$

$x^{\frac{2}{5}.5}(\log x{)}^5={10}^2$

$(x^{\frac{2}{5}}\log x{)}^5={10}^2$

$x^{\frac{2}{5}}\log x={10}^{\frac{2}{5}}$

$\log x^{x^{\frac{2}{5}}}={10}^{\frac{2}{5}}$

$x^{x^{\frac{2}{5}}}={10}^{{10}^{\frac{2}{5}}}$

Comparing the both side, we get

$x=10$