Log(1/13) (2x-1)+log(1/13) x>0

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To solve this logarithmic equation, we'll first combine the two logarithms using the product rule of logarithms:

log(1/13) [(2x - 1)x] = log(1/13) (2x^2 - x)

Now, we know that log(a) - log(b) = log(a/b), so we can rewrite the equation as:

log(1/13) (2x^2 - x) = log(1/13)

Since the logarithmic functions are equal, the expressions inside the logs must also be equal:

2x^2 - x = 1

Rearranging the equation, we have:

2x^2 - x - 1 = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = [-(-1) ± √((-1)^2 - 42(-1))]/(2*2
x = [1 ± √(1 + 8)]/
x = [1 ± √9]/
x = [1 ± 3]/4

Therefore, the two possible solutions for x are
1) x = (1 + 3)/4 = 4/4 =
2) x = (1 - 3)/4 = -2/4 = -1/2

Since x > 0, the solution is x = 1.