Question

If $S$ be a set defined as
$S=\{x\in R:{\log }_{5}120+(x-3)-2{\log }_5(1-5^{x-3})=-{\log }_5(0.2-5^{x-4}\left)\right\}$, then which of the following statement is true?

• A. $S$ is an empty set
• B. $S$ contains exactly one element
• C. $S$ contains exactly two elements
• D. $S$ contains exactly three elements

Answer 1

The correct option is B $S$ contains exactly one element

For the given equation to be defined, $1-5^{x-3}>0$ and $0.2-5^{x-4}>0$
$\Rightarrow 5^{x-3}<1$ and $5^{x-4}<0.2$
$\Rightarrow 5^{x-3}<1$ and $5^{x-4}<\frac{1}{5}$
$\Rightarrow x<3$ and $x<3$
$\therefore x<3\cdots \left(1\right)$

${\log }_{5}120+(x-3)-2{\log }_5(1-5^{x-3})=-{\log }_5(0.2-5^{x-4})$
$\Rightarrow {\log }_{5}120+{\log }_5{5}^{x-3}-{\log }_5(1-5^{x-3}{)}^2+{\log }_5(0.2-5^{x-4})=0$
$\Rightarrow {\log }_5\frac{120\times 5^{x-3}(0.2-5^{x-4})}{(1-5^{x-3}{)}^2}=0$
$\Rightarrow \frac{120\times 5^{x-3}(0.2-5^{x-4})}{(1-5^{x-3}{)}^2}=1$
$\Rightarrow \frac{120\times 5^{x-3}(1-5^{x-3})}{(1-5^{x-3}{)}^2\times 5}=1$
$\Rightarrow \frac{24\times 5^{x-3}(1-5^{x-3})}{(1-5^{x-3}{)}^2}=1$

Assuming $5^{x-3}=t$, we get
$\frac{24\times t(1-t)}{(1-t{)}^2}=1\Rightarrow 24t(1-t)=(1-t{)}^2\Rightarrow (1-t\left)\right[24t-(1-t)]=0\Rightarrow t=1\text{or}t=\frac{1}{25}\Rightarrow 5^{x-3}=1\text{or}{5}^{x-3}=\frac{1}{25}\Rightarrow x-3=0\text{or}x-3=-2\Rightarrow x=3\text{or}x=1$
As $x<3\Rightarrow x=1$
$\therefore S=\left\{1\right\}$