Question

If the third term in the expansion of ${\left(\frac{1}{x}+x^{lo{g}_{10^x}}\right)}^5$ is $1000$, then $x=$

• A. $10^2$
• B. $10^3$
• C. $10$
• D. $10^4$

Answer 1

Answer: (A)

Given, $T_3=1000$ in the expansion of
${\left(\frac{1}{x}+x^{lo{g}_{10^x}}\right)}^5$
$\therefore {}^5{C}_2{\left(\frac{1}{x}\right)}^3{\left(x^{lo{g}_{10^x}}\right)}^2=1000$
$\Rightarrow 10x^{-3}\times x^{2lo{g}_{10^x}}=1000$
$\Rightarrow x^{2lo{g}_{10^{x-3}}}=10^2$
$\Rightarrow \left(2lo{g}_{10}x-3\right)=lo{g}_{x}10^2$
(Taking logarithm on both sides to the base $x$)
$\left(2lo{g}_{10}x-3\right)=\frac{2}{lo{g}_{10}x}$
$\Rightarrow 2t-3=\frac{2}{t}$, where $t=lo{g}_{10}x$
$\Rightarrow \left(2t+1\right)\left(t-2\right)=0$
$\Rightarrow t=2$ ($t\ne -1/2$, neglected)
$\therefore lo{g}_{10}x=2$
$\Rightarrow x=10^2=100$