We need to find x if 5*6^x-2*3^2x-3*2^2x=0

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

divide all the terms by 2^2x

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

let (3/2)^x = y

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

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

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

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

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

=> y = 1 and y = 3/2

y = (3/2)^x

(3/2)^x = 1

=> x = 0

(3/2)^x = 3/2

=> x = 1

**The values of x are 1 and 0**

We'll begin by multiplying by -1 and we'll re-write the equation:

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

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

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

We'll divide by 3^2x:

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}.**