Log2x+log2(x-1)<_1

To solve the inequality \( \log_2(x) + \log_2(x-1) \leq 1 \), we can first use the properties of logarithms to combine the left-hand side: \[ \log_2(x) + \log_2(x-1) = \log_2(x(x-1)) \] This means the inequality can be rewritten as: \[ \log_2(x(x-1)) \leq 1 \] Next, we can exponentiate both sides to eliminate the logarithm (keeping in mind that since the base \( 2 \) is positive and greater than one, the inequality will remain unchanged): \[ x(x-1) \leq 2^1 \] This simplifies to: \[ x(x-1) \leq 2 \] Now, we can rearrange the inequality: \[ x^2 - x - 2 \leq 0 \] Next, we need to factor the quadratic: \[ (x - 2)(x + 1) \leq 0 \] Now we can find the critical points by setting the expression equal to zero: \[ (x - 2)(x + 1) = 0 \] This gives us the roots: \[ x = 2, \quad x = -1 \] We can now test the intervals determined by these roots: \( (-\infty, -1) \), \( (-1, 2) \), and \( (2, \infty) \). 1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \): \[ (-2 - 2)(-2 + 1) = (−4)(-1) = 4 > 0 \] 2. **Interval \( (-1, 2) \)**: Choose \( x = 0 \): \[ (0 - 2)(0 + 1) = (-2)(1) = -2 \leq 0 \] 3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \): \[ (3 - 2)(3 + 1) = (1)(4) = 4 > 0 \] Now, we summarize our results. The quadratic \( (x - 2)(x + 1) \leq 0 \) is satisfied in the interval \( [-1, 2] \). However, we must also consider the restrictions from the logarithmic expressions. Since \( \log_2(x) \) and \( \log_2(x-1) \) must be defined, we require: 1. \( x > 0 \) 2. \( x - 1 > 0 \) implies \( x > 1 \) Thus, combining these restrictions with the inequality solution, we get: \[ 1 < x \leq 2 \] So, the final solution to the inequality \( \log_2(x) + \log_2(x-1) \leq 1 \) is: \[ \boxed{(1, 2]} \]