Question

If the fourth term in the expansion of ${{\left( \sqrt{{{x}^{\dfrac{1}{\log x+1}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}$ is equal to $200$ and $x>1$ , then $x$ is
A. $10$
B. ${{10}^{-4}}$
C. $1$
D. $-4$

Answer 1

Hint: ${{T}_{4}}$ is given. Solve it as ${{T}_{3+1}}$ you will get the value of $x$. Use ${{T}_{3+1}}$ as ${{(r+1)}^{th}}$ term and solve it.

Complete step by step answer:
So from the above question we got to know that the value of ${{T}_{4}}$ is given i.e. $200$ . We have to find the value of $x$ . So for finding the value of $x$ We have to use the ${{(r+1)}^{th}}$ term which is given below.
So ${{T}_{4}}=200$ ……….. (1)
So if we take a ${{(a+b)}^{n}}$ ,
We know we get the value of ${{T}_{r+1}}$ ,
So the value of ${{T}_{r+1}}$ is as follows,
i.e ${{T}_{r+1}}{{=}^{n}}{{c}_{r}}{{a}^{n-r}}{{b}^{r}}$ …………. (So this is ${{(r+1)}^{th}}$ term)…….. (2)
So here $a=\sqrt{{{x}^{\dfrac{1}{\log x+1}}}}$ and $b={{x}^{\dfrac{1}{12}}}$ ………… (3)
So from (1) i.e. ${{T}_{4}}$ we can write it as,
${{T}_{3+1}}=200$ ….. (4)
Where we get $r=3,$
So we have to apply this ${{(r+1)}^{th}}$ term, So applying it we it as follows,
So from (2), (3) and (4), We get,
${{T}_{3+1}}{{=}^{6}}{{c}_{3}}{{\left[ \sqrt{{{x}^{\dfrac{1}{\log x+1}}}} \right]}^{(6-3)}}{{({{x}^{\dfrac{1}{12}}})}^{3}}=200$
So simplifying we get,
$\dfrac{6\times 5\times 4}{3\times 2\times 1}{{\left[ \sqrt{{{x}^{\dfrac{1}{\log x+1}}}} \right]}^{(3)}}({{x}^{\dfrac{1}{4}}})=200$
$(5\times 4){{\left[ \sqrt{{{x}^{\dfrac{1}{\log x+1}}}} \right]}^{(3)}}({{x}^{\dfrac{1}{4}}})=200$
So simplifying we get,
$(20){{\left( {{x}^{\dfrac{1}{\log x+1}}} \right)}^{\left( \dfrac{3}{2} \right)}}({{x}^{\dfrac{1}{4}}})=200$
Dividing above Whole equation by $20$ , We get,
${{\left( {{x}^{\dfrac{1}{\log x+1}}} \right)}^{\left( \dfrac{3}{2} \right)}}({{x}^{\dfrac{1}{4}}})=\dfrac{200}{20}$
${{\left( {{x}^{\dfrac{1}{\log x+1}}} \right)}^{\left( \dfrac{3}{2} \right)}}({{x}^{\dfrac{1}{4}}})=10$
So simplifying in simple manner we get,
$\left( {{x}^{\dfrac{3}{2(\log x+1)}}} \right)({{x}^{\dfrac{1}{4}}})=10$………. (5)
We know the property,
${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$
So we have to apply the above property,
So applying the property we get,
So using the above property in (5), We get,
$\left( {{x}^{\dfrac{3}{2(\log x+1)}+\dfrac{1}{4}}} \right)=10$
Also we know the property ${{a}^{{{\log }_{a}}6}}=6$ So applying same property we get,
$\left( {{x}^{\dfrac{3}{2(\log x+1)}+\dfrac{1}{4}}} \right)={{x}^{{{\log }_{x}}10}}=10$
So equating we get,
$\dfrac{3}{2(\log x+1)}+\dfrac{1}{4}={{\log }_{x}}10$
So writing ${{\log }_{x}}10=\dfrac{1}{{{\log }_{10}}x}$ .
Now let us substitute ${{\log }_{10}}x=y$,
So Substituting above we get,
$\dfrac{3}{2(y+1)}+\dfrac{1}{4}=\dfrac{1}{y}$
So simplifying it we get, i.e. taking LCM,
We get it as,
$\dfrac{3\times 2}{2\times 2(y+1)}+\dfrac{(y+1)}{4\times (y+1)}=\dfrac{1}{y}$
$\dfrac{6+y+1}{4(y+1)}=\dfrac{1}{y}$
So now cross multiplying,
we get it as follows,
$(6+y+1)\times y=4(y+1)$
$\begin{aligned} & 6y+{{y}^{2}}+y=4y+4 \\\ & {{y}^{2}}+7y-4y-4=0 \\\ & {{y}^{2}}+3y-4=0 \\\ & {{y}^{2}}+4y-y-4=0 \\\ & y(y+4)-(y+4)=0 \\\ & (y+4)(y-1)=0 \\\ \end{aligned}$
So we get from above the value of $y$ ,
$y=-4$ and $y=1$
We know that ${{\log }_{10}}x=y$,
So We also get $y=-4$,
$\begin{aligned} & {{\log }_{10}}x=-4 \\\ & x={{10}^{-4}} \\\ \end{aligned}$
For $y=1$
$\begin{aligned} & {{\log }_{10}}x=1 \\\ & x=10 \\\ \end{aligned}$
Since, $x>1$ , so the value of $x=10$ .

Option (A) is correct.

Note: Be familiar with the expansion and ${{T}_{r+1}}{{=}^{n}}{{c}_{r}}{{a}^{n-r}}{{b}^{r}}$ remember this ${{(r+1)}^{th}}$ term.
You should know the properties of log so you can solve the sum. The properties such as ${{\log }_{x}}10=\dfrac{1}{{{\log }_{10}}x}$ etc. Also be familiar with ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ this property. While solving, take care of conversions.