Question

The sum of all possible values of $x$ satisfying the equation $6{\log }_{(x-1)}10+{\log }_{10}(x-1)=5$ is ?

• A. $101$
• B. $1101$
• C. $1001$
• D. $1102$

Answer 1

Answer: (D)

$1102$

$6{\log }_{x-1}10+{\log }_{10}(x-1)=5$

$\Rightarrow \frac{6}{{\log }_{10}(x-1)}+{\log }_{10}(x-1)=5$ $\{\because {\log }_{b}a=({\log }_{a}b{)}^{-1}\}$

let ${\log }_{10}(x-1)=t$

$\therefore \frac{6}{t}+t=5$

$\Rightarrow t^2-5t+6=0$

$(t-2)(t-3)=0$

$\Rightarrow t=2,3$

$\Rightarrow {\log }_{10}(x-1)=2,3$

$x-1=100,1000$

$\Rightarrow x=101,1001$

Sum of values of $x=101+1001$

$=1102$