Question

$\log_{3}(6-2x) > \log_{3}(4+x)$

Answer 1

The solution set is $(-4, \frac{2}{3})$.

Answer 2

We are asked to solve the inequality $\log_{3}(6-2x) > \log_{3}(4+x)$.

For the logarithms to be defined, the arguments must be positive.

1. $6 - 2x > 0$
   $6 > 2x$
   $3 > x$
   $x < 3$

2. $4 + x > 0$
   $x > -4$

Combining these two conditions, the domain of the inequality is $-4 < x < 3$.

Since the base of the logarithm is $3$, which is greater than $1$, the inequality holds for the arguments in the same direction as the logarithms.
$\log_{3}(6-2x) > \log_{3}(4+x) \implies 6 - 2x > 4 + x$

Now, we solve this linear inequality:
$6 - 4 > x + 2x$
$2 > 3x$
$x < \frac{2}{3}$

The solution to the original inequality must satisfy both the domain condition ($-4 < x < 3$) and the condition derived from comparing the arguments ($x < \frac{2}{3}$). We need to find the intersection of the interval $(-4, 3)$ and the interval $(-\infty, \frac{2}{3})$.

The intersection is $(-4, 3) \cap (-\infty, \frac{2}{3})$.
On a number line, the interval $(-4, 3)$ is from -4 up to 3 (exclusive).
The interval $(-\infty, \frac{2}{3})$ is from negative infinity up to $\frac{2}{3}$ (exclusive).
The intersection of these two intervals is the set of numbers that are greater than -4 AND less than 3 AND less than $\frac{2}{3}$.
Since $\frac{2}{3} < 3$, the condition $x < 3$ is automatically satisfied if $x < \frac{2}{3}$.
Therefore, the intersection is $-4 < x < \frac{2}{3}$.

The solution set is the interval $(-4, \frac{2}{3})$.

Explanation:

1. Establish the domain by requiring the arguments of the logarithms to be positive: $6-2x > 0$ and $4+x > 0$. This yields the domain $-4 < x < 3$.
2. Since the base of the logarithm (3) is greater than 1, the inequality $\log_b a > \log_b c$ is equivalent to $a > c$. Apply this to the given inequality: $6-2x > 4+x$.
3. Solve the resulting linear inequality: $6-2x > 4+x \implies 2 > 3x \implies x < 2/3$.
4. The solution to the original inequality is the intersection of the domain and the solution from step 3: $(-4, 3) \cap (-\infty, 2/3) = (-4, 2/3)$.