Question

Log3(x+2)<1 А(1;бесконечность )Б(-2;бесконечность) В(-2;1) Г(-бесконеч;1)

Answer 1

$log_3(x+2)\ \textless \ 1\; \; ,\; \; \; ODZ:\; \; x+2\ \textgreater \ 0\; ,\; x\ \textgreater \ -2\\\\log_3(x+2)\ \textless \ log_33\\\\x+2\ \textless \ 3\\\\x\ \textless \ 1\\\\ \left \{ {{x\ \textgreater \ -2} \atop {x\ \textless \ 1}} \right. \; \; \; \Rightarrow \; \; \; \underline {x\in (-2,1)}$

Answer 2

Решение
log3(x+2)<1, x>-2
x+2<3
x<3-2
x<1, x>-2
ответ: (-2;1)

Answer 3

To solve the inequality $\log_3(x+2) < 1$, we need to find the values of $x$ that satisfy the inequality. Remember that $\log_3(x+2)$ represents the logarithm base 3 of the expression $x+2$.

Step 1: Rewrite the inequality to get rid of the logarithm: $\log_3(x+2) < 1$

Step 2: Exponentiate both sides using the base 3: $3^{\log_3(x+2)} < 3^1$

Step 3: Use the property $3^{\log_3(x+2)} = x+2$ on the left side: $x+2 < 3$

Step 4: Solve for $x$: $x < 3-2$ $x < 1$

So, the solution to the inequality is $x < 1$.

Now, let's interpret the options A, B, C, and D:

A: (1; infinity) - This means all real numbers greater than 1, but our solution is $x < 1$, so this option is incorrect.

B: (-2; infinity) - This means all real numbers greater than -2, but our solution is $x < 1$, so this option is also incorrect.

C: (-2; 1) - This means all real numbers between -2 and 1, excluding -2 and 1. This matches our solution $x < 1$, so this option is correct.

D: (-infinity; 1) - This means all real numbers less than 1, but our solution is $x < 1$, not $x \leq 1$, so this option is incorrect.

Therefore, the correct option is В(-2;1).