We have to solve 8*15^x/5 - 3*5^(2x-1) - 9^x = 0

8*15^x/5 - 3*5^(2x-1) - 9^x = 0

=> 8*15^x/5 - 3*5^2x/5 - 3^2^x = 0

=> (8/5)*3^x*5^x - (3/5)*5^x^2 - 3^x^2 = 0

divide by 5^2^x

=> (8/5)*(3/5)^x - (3/5) - (3/5)^x^2 = 0

Let (3/5)^x = y

=> (8/5)*y - (3/5) - y^2 = 0

=> 5y^2 - 8y + 3 = 0

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

=> 5y (y - 1) - 3(y - 1) = 0

=> (5y - 3)(y - 1) = 0

=> y = 3/5 and y = 1

But y = (3/5)^x

(3/5)^x = 3/5

=> x = 1

(3/5)^x = 1

=> x = 0

**The solutions of x are x = 1 and x = 0.**

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

3^2x - (8/5)*15^x + (3/5)*5^2x = 0

We remark that 15^x = (3*5)^x

But (3*5)^x = 3^x*5^x

We'll divide by 5^2x all over:

(3/5)^2x - (8/5)*(3/5)^x + 3/5 = 0

We'll note (3/5)^x = t

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

(3/5)^2x = t^2

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

t^2 - 8t/5 + 3/5 = 0

We'll notice that the sum of the roots is 8/5 and the product is 3/5.

3/5 + 1 = 8/5

3/5*1 = 3/5

The roots of the quadratic are t1 = 3/5 and t2 = 1.

Now, we'll put (3/5)^x = t1:

(3/5)^x = 3/5

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

x = 1

We'll write 1 = (3/5)^0

(3/5)^x = t2

(3/5)^x = (3/5)^0

x = 0

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