Question

Lg(x-1)<=0 help..........​

Answer 1

Ответ:

Пошаговое объяснение:

сначала запишем, что х-1 > 0 , это по определению логарифма, тогда мы имеем первое условие на х   x > 1

потом заменим 0 в левой части на lg(1) т.к. lg(1) =0

и получим

lg(x-1) ≤ lg(1), основание больше единицы, поэтому имеем

х -1 ≤ 1  ⇒  х ≤ 2

теперь соединим это с первым условием и получим решение

1 <  x ≤ 2 или х ∈ (1; 2]

Answer 2

To solve the inequality $lg(x-1) \leq 0$, you need to find the values of $x$ that satisfy it. In this case, $lg$ refers to the logarithm with base 10, commonly denoted as $\log_{10}$. Here's how you can solve it step by step:

1. Start by writing the inequality: $\log_{10}(x-1) \leq 0$

2. To eliminate the logarithm, you can rewrite it as an exponent. Remember that $\log_{10}(x-1)$ is asking, "What power must 10 be raised to in order to obtain $x-1$?" So, rewrite it as: $10^0 \leq x-1$

3. Since any non-zero number raised to the power of 0 is equal to 1 ($a^0 = 1$ for $a \neq 0$), the left side simplifies to 1: $1 \leq x-1$

4. Now, isolate $x$ by adding 1 to both sides of the inequality: $1 + 1 \leq x$

5. Simplify the right side: $2 \leq x$

So, the solution to the inequality $\log_{10}(x-1) \leq 0$ is $x \geq 2$. In interval notation, you can express this as $[2, \infty)$, meaning that $x$ is greater than or equal to 2.