Question

$3^{2x+1}=3^{x+2}+\sqrt{1-{6.3}^x+3^{2(x+1)}}$

Answer 1

$3^{2x+1}=3^{x+2}+\sqrt{1-{6.3}^x+3^{2(x+1)}}$

$\Rightarrow 3\times 3^{2x}=(9\times 3^x)+\sqrt{1-{6.3}^x+(9\times 3^{2x})}$

tke $3^x=t$

$\Rightarrow 3t^2=\left(9t\right)+\sqrt{1-6t+(9\times t^2)}$

$\Rightarrow 3t^2=9t+\sqrt{1-6t+9t^2}$

$\Rightarrow 3t^2=9t+\sqrt{(3t-1{)}^2}$

$\Rightarrow 3t^2=9t+(3t-1)$

$\Rightarrow 3t^2-12t+1=0$

On solving we get

$\Rightarrow t=\frac{6\pm \sqrt{33}}{3}$

$\Rightarrow 3^x=\frac{6\pm \sqrt{33}}{3}$

$\Rightarrow x={\log }_3\left(\frac{6\pm \sqrt{33}}{3}\right)$