Question

If the third term in the expansion of ${[x+x^{{\log }_{10}x}]}^5$ is ${10}^6$, then $x$ can be

• A. ${10}^{-1/3}$
• B. $10$
• C. ${10}^{-5/2}$
• D. ${10}^2$

Answer 1

Answer: (B), (C)

The correct options are
B $10$
C ${10}^{-5/2}$

$T_3={}^5{C}_2{x}^{3+2lo{g}_{10}x}$
$=10x^{lo{g}_{10}\left({10}^3{x}^2\right)}$
$={10}^6$
Hence
${10}^5=x^{{\log }_{10}\left({10}^3{x}^2\right)}$
Taking ${\log }_{10}$ on both sides, we get
$5={\log }_{10}\left({10}^3{x}^2\right)\log \left(x\right)$
$5=(3+2\log (x\left)\right)\log \left(x\right)$
$5=3\log \left(x\right)+2\log (x{)}^2$
$2\log (x{)}^2+3\log (x)-5=0$
$2\log (x{)}^2+5\log (x)-2\log (x)-5=0$
$\log \left(x\right)\left(2\log \right(x)+5)-1\left(2\log \right(x)+5)=0$
$\left(\log \right(x)-1)\left(2\log \right(x)+5)=0$
$\log \left(x\right)=1$
$x=10$ and $\log \left(x\right)=\frac{-5}{2}$
$x={10}^{\frac{-5}{2}}$