Question

Решить уравнение log7(9-x)>_log7 8

Answer 1

log₇(9-x)≥log₇8

ОДЗ: 9-x>0; -x>-9; x<9

9-x≥8

-x≥8-9

-x≥-1

x≤1

Ответ: x∈(-∞;1]

Answer 2

To solve the inequality $\log_7(9-x) \geq \log_7 8$, we need to find the range of values for $x$ that satisfy the inequality.

Step 1: Recall the logarithmic property that $\log_b(x) \geq \log_b(y)$ is equivalent to $x \geq y$, where $b$ is the base of the logarithm.

Step 2: Apply the property to the given inequality:

$\log_7(9-x) \geq \log_7 8$ is equivalent to $9 - x \geq 8$.

Step 3: Now, solve for $x$:

$9 - x \geq 8$

Subtract 9 from both sides:

$-x \geq -1$

Divide by -1 (since we are dividing by a negative number, the inequality sign flips):

$x \leq 1$

Step 4: The solution to the inequality is $x \leq 1$.

In interval notation, the solution is $(- \infty, 1]$. This means that any value of $x$ less than or equal to 1 will satisfy the original inequality.