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
