Question

If the 6th term in the expansion of
$( \frac{1}{ {x}^{ \frac{8}{3} } } + {x}^{2} log_{10}(x) {)}^{8}$
is 5600, then x equals
(1) 1
(3) 10
(2) log, 10
(4) r does not exist​

Answer 1

Question :

If the 6th term in the expansion of $\sf[\dfrac{1}{x^{\frac{8}{3}}}+x^2\:\log_{10}(x)]^8$ is 5600

then find the value of x .

Theory :

•Binomial Theorem

If x and a are real number , then for all n € N,

$\sf\:(x+a)^n=\sum\limits_{r = 0}^{n}\:\:^{n}C_r\:x^{n-r}\:a^r$

• Genral term

$\sf\:T_{r+1}=^{n}C_r\:x^{n-r}\:a^r$

Solution :

Given : 6th term in expansion of $\sf[\dfrac{1}{x^{\frac{8}{3}}}+x^2\:\log_{10}(x)]^8$ is 5600

We know that

$\sf\:T_{r+1}=^{n}C_r\:x^{n-r}\:a^r$

Thus , 6th term in the given binomial expression:

$\sf\implies\:T_6=T_{5+1}=^{8}C_5(\dfrac{1}{x^{\frac{8}{3}}})^3\:(x^2\:\log_{10}\:x)^5$

$\sf\implies\:T_6=^{8}C_5\:\dfrac{1}{x^8}\:x^{10}\:(\log_{10}\:x)^5$

$\sf\implies\:T_6=^{8}C_5\:\dfrac{x^{10}}{x^8}\:(\log_{10}\:x)^5$

$\sf\implies\:T_6=^{8}C_5\:x^2\:(\log_{10}\:x)^5$

We know that

$\sf\green{^{n}C_r=\dfrac{n\:!}{r\:!(n-r)\:!}}$

Then ,

$\sf\:T_6=\dfrac{8\:!}{5\:!\times3\:!}\:x^2\:(\log_{10}\:x)^5$

$\sf\implies\:5600=56\times\:x^2\:(\log_{10}\:x)^5$

$\sf\implies\:100=x^2\:(\log_{10}\:x)^5$

$\sf\implies\:10^2\:(\log_{10}\:10)^5=x^2\:(\log_{10}\:x)^5$

On Comparing :

$\rm\blue{x=10}$

Therefore , correct option is b) x= 10

Answer 2

Answer:

x=10 is the right answer