#log(x + 10) - log(3x - 8) = 2#
Recall the law of logarithm..
#loga - logb = loga/logb#
Hence;
#log(x + 10)/log(3x - 8) = 2 -> "applying law of logarithm"#
#log[(x + 10)/(3x - 8)] = 2#
For every log, there is a base, and it's default base is #10#, since there was no actual base specified in the given question..
Therefore;
#log_10 [(x + 10)/(3x - 8)] = 2#
Recall again the law of logarithm..
#log_a x = y rArr x = a^y#
Hence;
#(x + 10)/(3x - 8) = 10^2 -> "applying law of logarithm"#
#(x + 10)/(3x - 8) = 100#
#(x + 10)/(3x - 8) = 100/1#
#1(x + 10) = 100(3x - 8) -> "cross multiplying"#
#x + 10 = 300x - 800#
#300x - 800 = x + 10 -> "rearranging the equation"#
#300x - x = 10 + 800 ->"collecting like terms"#
#299x = 810#
#x = 810/299 -> "diving both sides by the coefficient of x"#
#x = 2.70#
