Question

1) log2(8-x)< 1 2) log2 x+log2(x-1) <либо=1

Answer 1

Это первый, второй аналогично можно решить

Answer 2

Let's solve the inequalities step by step:

1. $\log_2(8 - x) < 1$:

First, let's remove the logarithm by converting it to its exponential form: $2^{\log_2(8 - x)} < 2^1$ $8 - x < 2$ $-x < -6$ $x > 6$

So, the solution to the first inequality is $x > 6$.

2. $\log_2(x) + \log_2(x - 1) \leq 1$:

Using the properties of logarithms, we can combine the logarithms into a single logarithm: $\log_2(x(x - 1)) \leq 1$

Next, let's remove the logarithm: $x(x - 1) \leq 2^1$ $x^2 - x \leq 2$ $x^2 - x - 2 \leq 0$

Now, we can solve this quadratic inequality by factoring: $(x - 2)(x + 1) \leq 0$

This inequality holds when one of the factors is non-positive and the other is non-negative. That means: $x - 2 \leq 0 \quad \text{and} \quad x + 1 \geq 0$ $x \leq 2 \quad \text{and} \quad x \geq -1$

The solution is $-1 \leq x \leq 2$.

To summarize, the solutions to the inequalities are:

1. $x > 6$
2. $-1 \leq x \leq 2$