To solve this logarithmic equation, we can use the properties of logarithms to combine the two logarithmic terms on the left side of the equation.

Recall the property that states log_a(b) + log_a(c) = log_a(b * c)

Applying this property to the given equation, we get:
log_5[(2x-1)(4-x)] = 1

Rewrite the equation in exponential form to eliminate the logarithm:
5^1 = (2x-1)(4-x)

Solve the equation:
5 = 8x - 2 - x^2

Rearrange the equation into a quadratic form:
x^2 - 8x + 7 = 0

Factorize the quadratic equation:
(x - 7)(x - 1) = 0

Set each factor to zero:
x - 7 = 0 or x - 1 = 0

Solve for x:
x = 7 or x = 1

Therefore, the solutions to the equation log_5(2x-1) + log_5(4-x) = 1 are x = 7 and x = 1.