Question

1) log4 (2x-6)=log4 (x-3)2) log8 (5-x)>log8 (12-3x)3)log1/7 (2x-5)⩽log1/7 (x-6)​

Answer 1

Ответ:

3)

Пошаговое объяснение:

там легко очень прям

Answer 2

To solve these logarithmic inequalities, we'll use the properties of logarithms and algebraic techniques. Remember that when working with logarithmic inequalities, we need to be careful about the domain of the logarithmic expressions, as the logarithm of a non-positive number is undefined.

1. $\log_4(2x-6) = \log_4((x-3)^2)$

Using the property $\log_a(b) = \log_a(c)$ implies $b = c$, we have:

$2x - 6 = (x - 3)^2$

Now, expand the right side:

$2x - 6 = x^2 - 6x + 9$

Rearrange the equation:

$x^2 - 8x + 15 = 0$

Factor the quadratic equation:

$(x - 3)(x - 5) = 0$

This gives us two possible solutions: $x = 3$ or $x = 5$. However, we need to check that these solutions satisfy the domain of the original logarithmic expression.

For $x = 3$: The left-hand side of the original equation becomes $\log_4(2 \cdot 3 - 6) = \log_4(0)$, which is undefined. So, $x = 3$ is not a valid solution.

For $x = 5$: The left-hand side of the original equation becomes $\log_4(2 \cdot 5 - 6) = \log_4(4) = 1$. The right-hand side of the original equation becomes $\log_4((5 - 3)^2) = \log_4(4) = 1$.

Since both sides are equal, $x = 5$ is a valid solution.

2. $\log_8(5 - x) > \log_8(12 - 3x)$

Using the property $\log_a(b) > \log_a(c)$ implies $b > c$, we have:

$5 - x > 12 - 3x$

Simplify and solve for $x$:

$2x > 7$

$x > \frac{7}{2}$

3. $\log_{\frac{1}{7}}(2x - 5) \leq \log_{\frac{1}{7}}(x - 6)$

Using the property $\log_a(b) \leq \log_a(c)$ implies $b \leq c$, we have:

$2x - 5 \leq x - 6$

Simplify and solve for $x$:

$x \geq -1$

However, when dealing with logarithms in bases less than 1 (like $\frac{1}{7}$ in this case), we need to ensure that the arguments of the logarithms are positive.

For the given inequality: $2x - 5 > 0$ and $x - 6 > 0$

Solve these inequalities separately:

$2x > 5$ ⟹ $x > \frac{5}{2}$

$x > 6$

So, the valid solution is $x > 6$.

Combining the conditions, the final solution is $x > 6$.