Solve the equations 3^(3x-3)*5^(x-4)=15^2x/5^7

We notice that we can write 15 = 3*5 => 15^2x =
3^2x*5^2x

We'lldivide 15^2x by
5^7:

15^2x/5^7 =
3^2x*5^2x/5^7

We'll use qutotient property for the
exponentials that have matching bases:

3^2x*5^2x/5^7 =
3^2x*5^(2x-7)

We'll re-write the
equation:

3^(3x-3)*5^(x-4)=3^2x*5^(2x-7)

We'll
divide by 3^2x and we'll get:

3^(3x-3)*5^(x-4)/3^2x =
5^(2x-7)

We'll divide by
5^(x-4):

3^(3x-3)/3^2x =
5^(2x-7)/5^(x-4)

We'll subtract the
exponents:

3^(3x - 3 - 2x) = 5^(2x - 7 - x +
4)

We'll combine like terms inside
brackets:

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

We'll re-write the
equation:

3^x*3^-3 =
5^x*5^-3

3^x/3^3 =
5^x/5^3

We'll create matching bases. We'll divide by
5^x:

3^x/5^x*3^3 = 1/5^3

We'll
multiply by 3^3:

3^x/5^x =
3^3/5^3

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

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

x =
3

The real solution of the equation is x =
3.