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:

\[ \boxed{(1, 2]} \]