To solve this equation, we first need to rewrite it in exponential form.

Since the bases of the logarithms on both sides are 2, we know that if log₂(a) = log₂(b), then a = b.

Therefore, we have

x^2 + 7x - 5 = 4x - 1

Now, let's solve this quadratic equation.

x^2 + 7x - 5 = 4x - 1
x^2 + 3x - 4 = 0

Now we can factorize the quadratic equation

(x + 4)(x - 1) = 0

Setting each factor to zero:

x + 4 = 0
x = -4

x - 1 = 0
x = 1

Therefore, the solutions to the equation log(2)(x^2+7x-5) = log(2)(4x-1) are x = -4 and x = 1.