The equation to be solved is : log(3) x + 1/log(3) x = 5/2

let log(3) x = y

y + 1/y = 5/2

=> y^2 + 1 = 5y/2

=> 2y^2 - 5y + 2 = 0

=> 2y^2 - 4y - y + 2 = 0

=> 2y(y - 2) -1(y - 2) = 0

=> (2y - 1)(y - 2) = 0

y = 1/2 and y = 2

As y = log(3) x

log(3) x = 1/2

=> x = sqrt 3

log(3) x = 2

=> x = 3^2 = 9

**The solution of the equation is x = sqrt 3 and x = 9**

We'll multiply all terms both sides by the least common denominator.

2*log3 (x)*log3 (x) + 2 = 5*log3 (x)

2*[log3 (x)]^2 - 5*log3 (x) + 2 = 0

We'll replace log3 (x) by t.

2t^2 - 5t + 2 = 0

We'll apply quadratic formula:

t1 = [5+sqrt(25 - 16)]/4

t1 = (5+3)/4

t1 = 2

t2 = 1/2

But log3 (x)=t.

log3 (x)=t1 <=> log3 (x)=2

We'll take antilogarithms and we'll get:

x = 3^2

x = 9

log3 (x)=t2

x = sqrt 3

**Since both values are positive, we'll accept them as solutions of equation: {sqrt3 ; 9}.**