# Calculate the solutions of the equation (5^x)*(5^x)=11, if x*y=2(x-3)(y-3)+3.

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5^x * 5^x = 11

x*y = 2(x-3)(y-3) +3

Let us substitute with x= 5^x and y= 5^x

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

11= 2(5^x -3)^2 +3

8= 2(5^x -3)^2

4= (5^x -3)^2

==> 5^x1 -3 = 2 ==> 5^x1=5 ==> x1=1

==> 5^x2 -3 = -2 ==> 5^x2 = 1 ==> x2=0

Then the solution is:

x1=1 and x2= 0

We'll substitute x and y from the expression of the composition law by 5^x and we'll get:

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

But (5^x)*(5^x) = 11, so:

11 = 2(5^x - 3)^2 + 3

We'll expand the square:

11 = 2(5^2x - 6*5^x + 9) + 3

We'll oen the brackets and we'll move all terms to one side:

2*5^2x - 12*5^x + 21-11 = 0

We'll solve the exponential equation using the substitution technique:

5^x = t

2t^2 - 12t + 10 = 0

We'll apply the quadratic formula:

t1 = [12+sqrt(144-80)]/4

t1 = (12+8)/4

t1 = 5

t2 = (12-8)/4

t2 = 1

Now, we'll find out the x values:

5^x = 5

We'll use one to one property:

x = 1

5^x = 1

5^x = 5^0

x = 0

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

The given composition x*y = 2(x-3)(y-3)+3

To solve (5^x)*(5^x).

Solution:

Let 5^x = t.

(5^x)*(5^x) = 11 becomes

2(t-3)(t-3)+3 =11.

2(t-3)^2 = 11-3 =8.

(t-3)^2 =8/2 = 4.

t-3 =+srt2 or -2

t = 3+2 . Or t =3-2 = 1.

So 5^x =5 =5^1. Or 5^x = 1 = 5^0.

Therefore, x =1 or x = 0.