The equation to be solved is 5*6^x = 2*3^2x + 3*2^2x

5*6^x = 2*3^2x + 3*2^2x

=> 5 = 2*3^2x/6^x + 3*2^2x/6^x

=> 5 = 2*(3/2)^x + 3*(2/3)^x

let (3/2)^x = y

=> 5 = 2y + 3/y

=> 5y = 2y^2 + 3

=> 2y^2 - 5y + 3 = 0

=> 2y^2 - 2y - 3y + 3 = 0

=> 2y(y - 1) - 3(y - 1) = 0

=> (2y - 3)(y - 1) = 0

y = 3/2 and y = 1

y = (3/2)^x

(3/2)^x = 3/2 for x = 1

(3/2)^x = 1 for x = 0

**The solution for the equation is x = 0 and x = 1**

First, we'll multiply by -1 and we'll re-write the equation:

3*2^2x - 5*6^x + 2*3^2x = 0

We remark that 6^x = (2*3)^x

But (2*3)^x = 2^x*3^x

We'll divide by 3^2x all over:

3*(2/3)^2x - 5*(2/3)^x + 2 = 0

We'll note (2/3)^x = t

We'll square raise both sides and we'll get:

(2/3)^2x = t^2

We'll re-write the equation in the new variable t:

3t^2 - 5t + 2 = 0

We'll apply quadratic formula:

t1 = [5+sqrt(25-24)]/6

t1 = (5+1)/6

t1 = 1

t2 = (5-1)/6

t2 = 2/3

Now, we'll put (2/3)^x = t1:

(2/3)^x = 1

We'll write 1 = (2/3)^0

(2/3)^x = (2/3)^0

Since the bases are matching, we'll apply one to one property of exponentials:

**x = 0**

(2/3)^x = 2/3

**x = 1**

**The solutions of the equation are {0 ; 1}.**