We have to find x given 3*4^x + 2*9^x = 5*6^x.

3*4^x + 2*9^x = 5*6^x

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

Divide all the terms by 3^2x

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

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

Let y = (2/3)^x

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

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

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

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

y = 2/3 and y = 1

As y = (2/3)^x

x = 1 or x = 0

**Therefore x can be 0 and 1.**

We'll move all terms to one side:

3*4^x + 2*9^x - 5*6^x= 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 the exponential ratio (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, changing the variable:

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 = t2

(2/3)^x = 2/3

x = 1

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