# The value of x satisfying 2log_9(2(1/2)^x-1)=log_27((1/4)^x-4)^3 is?

Aug 1, 2018

$2 {\log}_{9} \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$

$\implies 2 \times {\log}_{9} \left(27\right) {\log}_{27} \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$

$\implies 2 \times {\log}_{9} \left({9}^{\frac{3}{2}}\right) {\log}_{27} \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$
$\implies 2 \times \frac{3}{2} {\log}_{9} \left(9\right) {\log}_{27} \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$

$\implies 3 {\log}_{27} \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$
$\implies {\log}_{27} {\left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right)}^{3} = {\log}_{27} {\left({\left(\frac{1}{4}\right)}^{x} - 4\right)}^{3}$

$\implies \left(2 {\left(\frac{1}{2}\right)}^{x} - 1\right) = \left({\left({\left(\frac{1}{2}\right)}^{x}\right)}^{2} - 4\right)$

Taking ${\left(\frac{1}{2}\right)}^{x} = y$ we get

$\left(2 y - 1\right) = \left({y}^{2} - 4\right)$

$\implies {y}^{2} - 2 y - 3 = 0$

$= \left(y + 1\right) \left(y - 3\right) = 0$

So ${\left(\frac{1}{2}\right)}^{x} = - 1 \to \text{not possible}$

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

$\implies - x \log 2 = \log 3$

$\implies x = - \log \frac{3}{\log} 2$