(4^x)*(2^x) = 12.....(1)

x*y = x+y +10........(2)

From (2), let us substitute x=4^x and y= 2^x

==> 4^x*2^x= 4^x+2^x +10

But from (1) we know that 4^x * 2^x=12

==>> 12= 4^x + 2^x +10

==> 2= 4^x + 2^x

==> 2= (2^x)^2 + 2^x

==> (2^x)^2+ 2^x -2= 0

Assume that y=2^x

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

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

==> y1= -2 =2^x (impossible)

==> y2= 1 = 2^X => x=0

First, we'll substitute x and y by 4^x and 2^x, in the expresison of the law of composition:

(4^x)*(2^x)= (4^x)+(2^x)+10

But, from enunciation, we'll have:

(4^x)*(2^x)=12

The equation will become:

12=(4^x)+(2^x)+10

We'll use the substitution technique to solve the exponential equation:

t^2+t+10-12=0

t^2+t-2=0

We'll apply the qudratic formula:

t1 = [-1+sqrt(1+8)]/2

t1 = (-1+3)/2

t1 = 1

t2 = -2

Now, we'll find out x values:

2^x = t1

2^x = 1

2^x = 2^0

We'll use one to one property:

x=0

2^x = t2

2^x = -2 impossible because 2^x>0!

**The only solution of the equation is x=0.**

To solve for (4^x)*(2^x) = 12 if x*y =x+y+10.

Under the composition (4^x)*(2^x) = 4^x+2^x+10 = 12 Or

4^x+2^x +10-12 = 0

(2^x)^2+2^x -2 = 0.

t^2+t-2 = 0.

(t-1)(t+2) = 0, wher t =2^x

Or

t =0 Or t+2 = 0

t = 0

2^x=1 implies x = 0.

t-2= 0 gives

2^x = -2 which does not give real root.