Question

If ${\log }_2\left(30\right)+(x-4)-2{\log }_2(1-2^{x-4})=-{\log }_2(0.5-2^{x-5})$, then the value of $x$ is

Answer 1

${\log }_2\left(30\right)+(x-4)-2{\log }_2(1-2^{x-4})=-{\log }_2(0.5-2^{x-5})$
$\Rightarrow {\log }_2(2\times 15)+(x-4)-2{\log }_2(1-2^{x-4})=-{\log }_2\left[0.5(1-2^{x-4})\right]$
$\Rightarrow 1+{\log }_2\left(15\right)+(x-4)-2{\log }_2(1-2^{x-4})=-{\log }_2\left(0.5\right)-{\log }_2(1-2^{x-4})$
$\Rightarrow 1+{\log }_2\left(15\right)+(x-4)=1+{\log }_2(1-2^{x-4})$
$\Rightarrow {\log }_2(1-2^{x-4})-{\log }_2\left(15\right)=x-4$
$\Rightarrow {\log }_2\left(\frac{1-2^{x-4}}{15}\right)=x-4$
$\Rightarrow 1-2^{x-4}=15\cdot 2^{x-4}$
$\Rightarrow 1=16\cdot 2^{x-4}$
$\Rightarrow 1=2^4\cdot 2^{x-4}$
$\Rightarrow 2^0=2^x$
$\Rightarrow x=0$