Question

Assertion :The number of solutions of the equation
${|x-3|}^{{\log }_2{x}^2-3{\log }_{x}4}=\frac{1}{x-3}$ is $4$ Reason: A polynomial equation of degree $n$ with real coefficients cannot have more than $n$ real roots.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

${|x-3|}^{{\log }_2{x}^2-3{\log }_{x}4}=\frac{1}{x-3}$ ....(1)
Equation (1) is valid when $x>0,x\ne 1,3$
Taking Logarithm with base 2 to equation (1), we get
$({\log }_2{x}^2-3{\log }_{x}4){\log }_2|x-3|=-{\log }_2(x-3)$
For $x>3$, $(2{\log }_{2}x-\frac{6}{{\log }_{2}x}){\log }_2(x-3)=-{\log }_2(x-3)$
$\Rightarrow {\log }_2(x-3)=0$ and $2{\log }_{2}x-\frac{6}{{\log }_{2}x}=-1$
$\Rightarrow x-3=1$ and $2{\left({\log }_{2}x\right)}^2+{\log }_{2}x-6=0$
$\Rightarrow x=4$ and ${\log }_{2}x=\frac{3}{2},-2$
Therefore, $x=\{4,2\sqrt{2},\frac{1}{4}\}$
Ans: D