To solve the equation [tex]\(\log_3(3x + 2) = \log_3(4x - 6)\)[/tex], we need to recognize that if two logarithms with the same base are equal, then their arguments must also be equal. Therefore, we can set the arguments of the logarithms equal to each other:
[tex]\[
3x + 2 = 4x - 6
\][/tex]
Next, we solve for [tex]\(x\)[/tex] by isolating the variable on one side of the equation. Start by subtracting [tex]\(3x\)[/tex] from both sides:
[tex]\[
3x + 2 - 3x = 4x - 6 - 3x
\][/tex]
This simplifies to:
[tex]\[
2 = x - 6
\][/tex]
Now, add 6 to both sides to isolate [tex]\(x\)[/tex]:
[tex]\[
2 + 6 = x - 6 + 6
\][/tex]
Simplifying that, we obtain:
[tex]\[
8 = x
\][/tex]
So, the solution to the equation [tex]\(\log_3(3x + 2) = \log_3(4x - 6)\)[/tex] is:
[tex]\[
x = 8
\][/tex]
Thus, the correct answer is [tex]\( \boxed{8} \)[/tex].
