Question

Решите неравенство 2lgx>1

Answer 1

Ответ:

2lg x> 1 или    lg (x^2)>  lg10,    x^2 >10.

Т.к. одз неравенства  x >0То ответ     x >√10.

Пошаговое объяснение:

Answer 2

Ответ:

$x\in(\sqrt{10}; +\infty)$

Пошаговое объяснение:

2lgx>1

Делим обе части на 2.

lgx>0,5

lgx>lg 10*0,5

lgx>*0,5*lg 10

По свойству логарифмов

a*lgb=lgbᵃ

$\lg x>\lg 10^{0,5}$

$\lg x>\lg\sqrt{ 10}$

Так как у десятичного логарифма lg x основание 10 больше 1, то потенцируем неравенство без изменения знака

$x>\sqrt{10}$

$x\in(\sqrt{10};+\infty)$

Answer 3

To solve the inequality $2\log(x) > 1$, we need to follow these steps:

Step 1: Rewrite the inequality in exponential form. Step 2: Solve for $x$.

Let's go through each step:

Step 1: Rewrite the inequality in exponential form. The logarithmic inequality $2\log(x) > 1$ can be rewritten using the logarithm property: $\log_b(a) > c$ is equivalent to $a > b^c$. In our case, $b = 10$ (since it's a common logarithm, base 10). So, we get:

$2\log(x) > 1 \implies \log(x) > \frac{1}{2}$

Step 2: Solve for $x$. Now, to remove the logarithm, we can rewrite the inequality in exponential form:

$x > 10^{1/2} = \sqrt{10} \approx 3.162$

So, the solution to the inequality $2\log(x) > 1$ is $x > 3.162$.