Question

The value of $x$ which satisfies the equation ${\log }_{\frac{1}{3}}(2{\left(\frac{1}{2}\right)}^x-1)={\log }_{\frac{1}{3}}({\left(\frac{1}{4}\right)}^x-4)$ is

• A. ${\log }_{2}3$
• B. $-{\log }_{2}3$
• C. $\frac{1}{{\log }_{3}2}$
• D. none of these.

Answer 1

The correct option is B $-{\log }_{2}3$

$\left(\right(\frac{1}{4}{)}^x-4)>0\Rightarrow x<-1$

The given equation to
$\{\begin{array}{c}2{\left(\frac{1}{2}\right)}^x-1>0\\ 2{\left(\frac{1}{2}\right)}^x-1={\left(\frac{1}{4}\right)}^x-4\end{array}$
$\Rightarrow \{\begin{array}{c}{\left(\frac{1}{2}\right)}^x>\frac{1}{2}\\ {\left(\frac{1}{2}\right)}^{2x}-2{\left(\frac{1}{2}\right)}^x-3=0\end{array}$
$\Rightarrow \{\begin{array}{c}x<1\\ [{\left(\frac{1}{2}\right)}^x-3][{\left(\frac{1}{2}\right)}^x+1]=0\end{array}$
$\Rightarrow \{\begin{array}{c}x<1\\ x=-{\log }_{2}3\end{array}$

Hence, $x=-{\log }_{2}3$ is the root of the original equation.