The product of all the solutions of the equation log x(3) ⋅ log (x/3)(3)+ log (x/81)(3)=0 is

Question

The product of all the solutions of the equation log x(3) ⋅ log (x/3)(3)+ log (x/81)(3)=0 is (A) 1 (B) 2 (C) 18 (D) 81

Tardigrade Admin · Accepted Answer

(1/ log 3 x ) ⋅ (1/( log 3 x -1))+(1/( log 3 x -4))=0 text Put log 3 x = t ( t -4)+ t ( t -1)=0 t 2=4 ⇒ t = ± 2 log 3 x =2 ⇒ x =9 log 3 x =-2 ⇒ x =(1/9)

Q. The product of all the solutions of the equation $lo g_{x} (3) \cdot lo g_{\frac{x}{3}} (3) + lo g_{\frac{x}{81}} (3) = 0$ is

1

2

18

81

$\frac{1}{l o g _{3} x} \cdot \frac{1}{( l o g _{3} x - 1 )} + \frac{1}{( l o g _{3} x - 4 )} = 0$
$Put lo g_{3} x = t$
$(t - 4) + t (t - 1) = 0$
$t^{2} = 4 \Rightarrow t = \pm 2$
$lo g_{3} x = 2 \Rightarrow x = 9$
$lo g_{3} x = - 2 \Rightarrow x = \frac{1}{9}$

Q. The product of all the solutions of the equation logx​(3)⋅log3x​​(3)+log81x​​(3)=0 is

1

2

18

81

log3​x1​⋅(log3​x−1)1​+(log3​x−4)1​=0 Put log3​x=t (t−4)+t(t−1)=0 t2=4⇒t=±2 log3​x=2⇒x=9 log3​x=−2⇒x=91​

Q. The product of all the solutions of the equation $lo g_{x} (3) \cdot lo g_{\frac{x}{3}} (3) + lo g_{\frac{x}{81}} (3) = 0$ is

$\frac{1}{l o g _{3} x} \cdot \frac{1}{( l o g _{3} x - 1 )} + \frac{1}{( l o g _{3} x - 4 )} = 0$
$Put lo g_{3} x = t$
$(t - 4) + t (t - 1) = 0$
$t^{2} = 4 \Rightarrow t = \pm 2$
$lo g_{3} x = 2 \Rightarrow x = 9$
$lo g_{3} x = - 2 \Rightarrow x = \frac{1}{9}$