Question

Logx (x-3) - logx (9-x)/ logx-1 X < 0

Answer 1

Ответ:

-x+(-(9-x)+(x-3) log (x)) <0

Пошаговое объяснение:

Answer 2

To solve the inequality $\log_x (x-3) - \log_x (9-x) < 0$ with the constraint $\log_{x-1} x > 0$, we need to follow a step-by-step process:

Step 1: Simplify the logarithmic expressions. Step 2: Combine the logarithms using logarithmic rules. Step 3: Solve for the intervals where the inequality holds true. Step 4: Apply the constraint $\log_{x-1} x > 0$ to find the final solution.

Let's go through each step:

Step 1: Simplify the logarithmic expressions. We know that $\log_a b - \log_a c = \log_a \frac{b}{c}$, so applying this rule: $\log_x \frac{x-3}{9-x} < 0$.

Step 2: Solve the inequality $\log_x \frac{x-3}{9-x} < 0$. The inequality is satisfied when the fraction is less than 1 but greater than 0: $0 < \frac{x-3}{9-x} < 1$.

Step 3: Solve for the intervals where the inequality holds true. Let's solve the individual inequalities: $0 < \frac{x-3}{9-x}$ and $\frac{x-3}{9-x} < 1$.

For the first inequality ($0 < \frac{x-3}{9-x}$): Cross-multiplying (remembering to consider the sign of the denominator): $0(9-x) < (x-3)(9-x)$, $0 < (x-3)(x-9)$.

This holds true when $3 < x < 9$.

For the second inequality ($\frac{x-3}{9-x} < 1$): Cross-multiplying (again considering the sign of the denominator): $(x-3)(9-x) < 9-x$, $-(x-3)^2 < 9-x$, $-x^2 + 6x - 9 < 9-x$, $-x^2 + 7x - 18 < 0$.

This quadratic expression factors as $-(x-9)(x+2) < 0$, which holds true when $-2 < x < 9$.

Step 4: Apply the constraint $\log_{x-1} x > 0$. Since $\log_{x-1} x$ is only defined when $x > 1$, we can narrow down our solution to the intersection of the intervals we found and the valid domain of $x$: $3 < x < 9$ and $1 < x < 9$, which simplifies to $3 < x < 9$.

So, the solution for the given inequality with the constraint is $3 < x < 9$.