# Find the value of x in  sqrt(1+log_x(27))*log_3(x)+1=0?

Jul 17, 2018

$\sqrt{1 + {\log}_{x} \left(27\right)} \cdot {\log}_{3} \left(x\right) + 1 = 0$
$\implies \sqrt{{\log}_{3} x \cdot {\log}_{3} x + 3 {\log}_{x} \left(3\right) {\log}_{3} \left(x\right) {\log}_{3} x} = - 1$

$\implies {\log}_{3} x \cdot {\log}_{3} x + 3 {\log}_{x} \left(3\right) {\log}_{3} \left(x\right) {\log}_{3} x = 1$

$\implies {\left({\log}_{3} x\right)}^{2} + 3 {\log}_{3} x - 1 = 0$

$\implies {y}^{2} + 3 y - 1 = 0$

$\implies y = \frac{- 3 \pm \sqrt{9 + 4}}{2}$

$\implies {\log}_{3} x = \frac{- 3 \pm \sqrt{9 + 4}}{2}$

So ${\log}_{3} x = 0.3028$

$\implies x = {3}^{0.3028} \approx 1.4$

Again
${\log}_{3} x = - 3.3$

$\implies x = {3}^{- 3.3} \approx 0.0265$