Express K in terms of L ?? #log_3 K - log_9 L = 2#

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)#