A bag contains 3 red balls and 4 black balls. If two balls are drawn without replacement, what is the probability that they are of different colors?

Number of red balls = 3
Number of black balls = 4
Total number of balls = $3 + 4 = 7$

There are two scenarios for selecting two different balls:

1. First ball is Red, second ball is Black (RB).
2. First ball is Black, second ball is Red (BR).

The probability of selecting a red ball first is $\frac{3}{7}$.
Since the first ball is not replaced, there are now 6 balls left (2 red, 4 black).
The probability of selecting a black ball second is $\frac{4}{6}$.

The probability of selecting a black ball first is $\frac{4}{7}$.
Since the first ball is not replaced, there are now 6 balls left (3 red, 3 black).
The probability of selecting a red ball second is $\frac{3}{6}$.

The probability of selecting two different balls is the sum of the probabilities of these two mutually exclusive events.

The probability of selecting two different balls is $\boxed{\frac{4}{7}}$.

By the definition of logarithms, if $\log_b a = c$, then $b^c = a$.
Applying this definition:

Take the square root of both sides:

For $\log_x A$ to be defined, the base $x$ must be positive and $x \neq 1$.
Therefore, $x = -\frac{3}{2}$ is not a valid solution.
The only valid solution is $x = \frac{3}{2}$.

The value of $x$ is $\boxed{\frac{3}{2}}$.