Question

The number of real value(s) of $x$ satisfying ${\log }_{1/3}(2{\left(\frac{1}{2}\right)}^x-1)={\log }_{1/3}({\left(\frac{1}{4}\right)}^x-4)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

${\log }_{1/3}(2{\left(\frac{1}{2}\right)}^x-1)={\log }_{1/3}({\left(\frac{1}{4}\right)}^x-4)$
For the $\log$ to be defined, we get
$2{\left(\frac{1}{2}\right)}^x-1>0\text{and}{\left(\frac{1}{4}\right)}^x-4>0\Rightarrow {\left(\frac{1}{2}\right)}^x>\frac{1}{2}\text{and}{\left(\frac{1}{4}\right)}^x>4\Rightarrow {\left(\frac{1}{2}\right)}^x>{\left(\frac{1}{2}\right)}^1\text{and}{\left(\frac{1}{4}\right)}^x>{\left(\frac{1}{4}\right)}^{-1}\Rightarrow x<1\text{and}x<-1(\because \frac{1}{4},\frac{1}{2}<1)\Rightarrow x<-1\cdots \left(1\right)$

${\log }_{1/3}(2{\left(\frac{1}{2}\right)}^x-1)={\log }_{1/3}({\left(\frac{1}{4}\right)}^x-4)\Rightarrow 2{\left(\frac{1}{2}\right)}^x-1={\left(\frac{1}{4}\right)}^x-4\Rightarrow 2{\left(\frac{1}{2}\right)}^x-1={\left(\frac{1}{2}\right)}^{2x}-4\Rightarrow {\left(\frac{1}{2}\right)}^{2x}-2{\left(\frac{1}{2}\right)}^x-3=0$

Assuming ${\left(\frac{1}{2}\right)}^x=t$
$\Rightarrow t^2-2t-3=0\Rightarrow (t+1)(t-3)=0\Rightarrow t=-1,3\Rightarrow {\left(\frac{1}{2}\right)}^x=3[\because {\left(\frac{1}{2}\right)}^x>0]$

Taking $\log$ on both sides, we get
$\Rightarrow x\log \frac{1}{2}=\log 3\Rightarrow x=-\frac{\log 3}{\log 2}\therefore x=-{\log }_{2}3$