Solve for x : 8*15^x/5 - 3*5^(2x-1) - 9^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}.