Find the value of x in $\sqrt{1 + {log}_{x} (27)} \cdot {log}_{3} (x) + 1 = 0$ $sqrt(1+log_x(27))*log_3(x)+1=0$ ?

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Find the value of x in $\sqrt{1 + {log}_{x} (27)} \cdot {log}_{3} (x) + 1 = 0$ $sqrt(1+log_x(27))*log_3(x)+1=0$ ?

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Answer 1 · 2018-07-17T16:17:10

$\sqrt{1 + {log}_{x} (27)} \cdot {log}_{3} (x) + 1 = 0$ $sqrt(1+log_x(27))*log_3(x)+1=0$
$\Rightarrow \sqrt{{log}_{3} x \cdot {log}_{3} x + 3 {log}_{x} (3) {log}_{3} (x) {log}_{3} x} = - 1$ $=>sqrt(log_3x*log_3x+3log_x(3)log_3(x)log_3x)=-1$

$\Rightarrow {log}_{3} x \cdot {log}_{3} x + 3 {log}_{x} (3) {log}_{3} (x) {log}_{3} x = 1$ $=>log_3x*log_3x+3log_x(3)log_3(x)log_3x=1$

$\Rightarrow {({log}_{3} x)}^{2} + 3 {log}_{3} x - 1 = 0$ $=>(log_3x)^2+3log_3x-1=0$

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

$\Rightarrow y = \frac{- 3 \pm \sqrt{9 + 4}}{2}$ $=>y=(-3pmsqrt(9+4))/2$

$\Rightarrow {log}_{3} x = \frac{- 3 \pm \sqrt{9 + 4}}{2}$ $=>log_3x=(-3pmsqrt(9+4))/2$

So ${log}_{3} x = 0.3028$ $log_3x=0.3028$

$\Rightarrow x = 3^{0.3028} \approx 1.4$ $=>x=3^(0.3028)~~1.4$

Again
${log}_{3} x = - 3.3$ $log_3x=-3.3$

$\Rightarrow x = 3^{- 3.3} \approx 0.0265$ $=>x=3^(-3.3)~~0.0265$

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Find the value of x in $\sqrt{1 + {log}_{x} (27)} \cdot {log}_{3} (x) + 1 = 0$ $sqrt(1+log_x(27))*log_3(x)+1=0$ ?

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