Question

If the third term in the expansion of $(\frac{1}{x}+x^{{\log }_{10}x}{)}^5$ is $1,000$, then $x$-equals

• A. 10${}^2$
• B. 10${}^3$
• C. $10$
• D. None of these

Answer 1

The correct option is A 10${}^2$

$T_{2+1}$
$=T_3$
$={}^5{C}_2{x}^{-3}{x}^{2lo{g}_{10}\left(x\right)}$
$={10}^3$
Therefore
$x^{-3}{x}^{2lo{g}_{10}\left(x\right)}=100$
$x^{2lo{g}_{10}\left(x\right)-3}=100$
Now x has to be in powers of 10.
Therefore
Let $x={10}^k$
Above equation reduces to
${10}^{k(2k-3)}={10}^2$
$2k^2-3k-2=0$
$(2k+1)(k-2)=0$
$k=2$ and $k=\frac{-1}{2}$
Hence $x={10}^2$ and $x={10}^{-\frac{1}{2}}$