Question

If $|x-1{|}^{\left(\right({\log }_{10}x{)}^2-{\log }_{10}{x}^2)}=|x-1{|}^3$, then

• A. $x=\frac{1}{10}$ is a solution for the equation
• B. $x=2$ is a solution for the equation
• C. Number of values of $x$ satisfying the equation is $4$
• D. $x=1$ is a solution for the equation

Answer 1

Answer: (A), (B)

The correct options are
A $x=\frac{1}{10}$ is a solution for the equation
B $x=2$ is a solution for the equation

$|x-1{|}^{\left(\right({\log }_{10}x{)}^2-{\log }_{10}{x}^2)}=|x-1{|}^3$
For the log to be defined,
$x>0$
If $a^m=a^3$
Then
$\left(i\right)m=3({\log }_{10}x{)}^2-{\log }_{10}{x}^2=3$
Assuming $t={\log }_{10}x$
$\Rightarrow t^2-2t-3=0$
$\Rightarrow (t-3)(t+1)=0$
$\Rightarrow {\log }_{10}x=-1,3$
$\Rightarrow x={10}^{-1},{10}^3$

$\left(ii\right)a=1|x-1|=1\Rightarrow x=0,2\Rightarrow x=2(\because x>0)$

$\left(iii\right)a=0,m\ne 0$
$a=0\Rightarrow |x-1|=0\Rightarrow x=1$
Checking the value of $m$ at $x=1$,
$m=({\log }_{10}1{)}^2-{\log }_{10}{1}^2=0$
Which contradict our assumption

Therefore the required values are
$x=\frac{1}{10},2,{10}^3$