To solve this logarithmic equation, we can use the properties of logarithms to combine the two logarithms into one. The properties we can use are:

log_a(b) + log_a(c) = log_a(bc)log_a(b) = c can be rewritten as a^c = b

Applying these properties to the given equation, we get:

log5((3x-1)*(3x-5)) = 1

Now, we can rewrite the equation in exponential form:

5^1 = (3x-1)*(3x-5)

5 = (3x-1)(3x-5)

5 = 9x^2 - 24x + 5

Rearranging the equation, we get:

9x^2 - 24x = 0

Factoring out 3x, we get:

3x(3x - 8) = 0

Setting each factor equal to zero, we solve for x:

3x = 0 => x = 0

3x - 8 = 0 => x = 8/3

Therefore, the solutions to the logarithmic equation log5(3x-1)+log5(3x-5)=1 are x = 0 and x = 8/3.