I shall do this step by step, using pka's idea, slightly modified.
But first here are the basic laws of logarithms.
Given that a, b, and c are positive real numbers, and b is not equal to 1:
[MATH]log_b(a) = x \iff a = b^x[/MATH].
[MATH]\therefore log_b(1) = 0 \text { and } log_b(b) = 1.[/MATH]
[MATH]log_b(a * c) = log_b(a) + log_b(c).[/MATH]
[MATH]-\ log_b(a) = log_b \left ( \dfrac{1}{a} \right ) \implies log_b(c) - log_b(a) = log_b \left ( \dfrac{c}{a} \right ).[/MATH]
[MATH]u log_b(a) = log_b(a^u).[/MATH]
[MATH]log_b(a) = log_b(c) \iff a = c.[/MATH]
These along with the change of base law must be memorized and understood to use logs.
Now your problem step by step the easy way
[MATH]log_b(x) = \frac{2}{3} log_b(8) + \frac{1}{2} log_b(9) - log(6)[/MATH].
[MATH]\therefore log_b(x) = \frac{2}{3} log_b(2^3) + \frac{1}{2} log_b(3^2) - log_b(6)[/MATH].
[MATH]\therefore log_b(x) =log_b(2^{(3 * 2 /3)})+ log_b(3^{(2 * 1/2)}) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(2^2) + log_b(3^1) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(4) + log_b(3) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b(4 * 3) - log_b(6) = log_b(12) - log_b(6).[/MATH]
[MATH]\therefore log_b(x) = log_b \left ( \dfrac{12}{6} \right ).[/MATH]
[MATH]\therefore log_b(x) = log_b(2).[/MATH]
[MATH]\therefore x = 2.[/MATH]
Learn the laws and be careful, and logs will give you no trouble.