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 - x^2 + 3x - 4 = 0
Now we can factorize the quadratic equation
(x + 4)(x - 1) = 0
Setting each factor to zero:
x + 4 = x = -4
x - 1 = x = 1
Therefore, the solutions to the equation log(2)(x^2+7x-5) = log(2)(4x-1) are x = -4 and x = 1.
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 -
x^2 + 3x - 4 = 0
Now we can factorize the quadratic equation
(x + 4)(x - 1) = 0
Setting each factor to zero:
x + 4 =
x = -4
x - 1 =
x = 1
Therefore, the solutions to the equation log(2)(x^2+7x-5) = log(2)(4x-1) are x = -4 and x = 1.