We can use a number of different properties of logarithms in working this problem:
log_a b - log_a c = log_a {:b/c:}color(white)("aaaaa")"Division Property"
log_a b = (log_c b)/(log_c a)color(white)("aaaaa")"Change of Base Property"
log_a b^c = c*log_a bcolor(white)("aaaaa")"Exponent Property"
log_a a = 1color(white)("aaaaa")"Identity Property"
a^(log_a b) = bcolor(white)("aaaaa")"Inverse Property"
We begin by rewriting the second logarithm as a base 3 logarithm using the Change of Base Property:
log_3 K - (log_3 L)/(log_3 9) = 2
Next, we apply the Exponent Property, followed by the Identity Property, on the second term:
log_3 K - (log_3 L)/(log_3 3^2) = 2
log_3 K - (log_3 L)/(2*log_3 3) = 2
log_3 K - (log_3 L)/(2*1) = 2
log_3 K - 1/2log_3 L = 2
Now we can use the reverse of the Exponent Property, and then recognize that an exponent of 1/2 represents a square root:
log_3 K - log_3 L^(1/2) = 2
log_3 K - log_3 sqrt(L) = 2
We can then use the Division Property, and finally remove the logarithms by using the definition of a logarithm:
log_3 {:K/sqrt(L):} = 2
3^(log_3 {:K/sqrt(L):}) = 3^2
K/sqrt(L) = 9
K = 9sqrt(L)