Question

If the fourth term in the expansion of ${(\sqrt{x^{\left(\frac{1}{logx+1}\right)}}+x^{\frac{1}{12}})}^6$ is equal to 200 and x > 1, then x is equal to

• A. ${10}^{\sqrt{2}}$
• B. $10$
• C. ${10}^4$
• D. None of these

Answer 1

Answer: (B)

$10$

${(a+b)}^6={{}^{6}C}_0{a}^6+{{}^{6}C}_1{a}^{5}b+{{}^{6}C}_2{a}^4{b}^2+{{}^{6}C}_3{a}^5{b}^3..........{{}^{6}C}_6{b}^6$

$\hookrightarrow$ 4th term

$t_4=200$

$\Rightarrow {{}^{6}C}_3{\left(\sqrt{x^{1/㏒x+1}}\right)}^3{\left(x^{1/12}\right)}^3=200$

$\Rightarrow 20\times \left[x^{3/2(logx+1)}\right]\left[x^{1/4}\right]=200$

$\Rightarrow x^{[3/2(logx+1)+1/4]}=10$

Taking log to base $10$ on both sides:

$[\frac{3}{2(logx+1)}+\frac{1}{4}]{log}_{10}x$ = ${log}_{10}10$

$\Rightarrow$ $\left[\frac{6+(1+logx)}{4(1+logx)}\right]logx=1$

$\Rightarrow (7+logx)logx=4+4logx$

$\Rightarrow {log}^{2}x+7logx=4+4logx$

$\Rightarrow {log}^{2}x+3logx-4=0$

$\Rightarrow (logx+4)(logx-1)=0$

$\Rightarrow x={10}^{-4}$ or $x={10}^1$

Given that, $x>1$

$\Rightarrow$ $x=10$

Hence the answer is $10.$