Question

Solve the following equation:
${\log }_{1/3}(2{\left(\frac{1}{2}\right)}^x-1)={\log }_{1/3}({\left(\frac{1}{4}\right)}^x-4)$

Answer 1

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

${\left(\frac{1}{2}\right)}^x$ cannot be negative.

Therefore, ${\left(\frac{1}{2}\right)}^x=3$

$x={-\log }_{2}3$