Question

Consider two sets defined as
$A=\{x:x\in R\text{and}\log (-x)=2\log (x+3\left)\right\},$
$B=\{x:x\in R\text{and}{x}^2+7x+9=0\}.$
Then which of the following is (are) TRUE?

• A. $n\left(A\right)=n\left(B\right)$
• B. $n(A\times B)=2$
• C. ${\log }_{n\left(B\right)}n\left(A\right)=0$
• D. ${\log }_{n\left(A\right)}n\left(B\right)$ is not defined

Answer 1

Answer: (B), (C), (D)

The correct options are
B $n(A\times B)=2$
C ${\log }_{n\left(B\right)}n\left(A\right)=0$
D ${\log }_{n\left(A\right)}n\left(B\right)$ is not defined

Equation $\log (-x)=2\log (x+3)$ is meaningful when
$-x>0$ and $x+3>0$
$\Rightarrow -3<x<0$

Now, $\log (-x)=2\log (x+3)$
$\Rightarrow -x=(x+3{)}^2$
$\Rightarrow x^2+7x+9=0\Rightarrow x=\frac{-7\pm \sqrt{13}}{2}$
Only $\frac{-7+\sqrt{13}}{2}\in (-3,0)$
$\therefore x=\frac{-7+\sqrt{13}}{2}$ is the only solution.
$\therefore A=\left\{\frac{-7+\sqrt{13}}{2}\right\}$
and $B=\left\{\frac{-7\pm \sqrt{13}}{2}\right\}$
$\Rightarrow n\left(A\right)=1,n\left(B\right)=2$

Clearly, $n\left(A\right)\ne n\left(B\right)$
$n(A\times B)=n\left(A\right)\cdot n\left(B\right)=2$
${\log }_{n\left(B\right)}n\left(A\right)={\log }_{2}1=0$
${\log }_{n\left(A\right)}n\left(B\right)={\log }_{1}2$ which is not defined as base is $1.$