Question 450128
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Hi
*[tex \LARGE\ \ log_bx - log_by = log_b(x/y) ]

1) log(2)xy-log(2)x^2 = log(2) xy/x^2 = log(2) y/x

In general, following true for logarithms
*[tex \large \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
*[tex \large\ \ nlog_bx = log_b(x^n) ]
*[tex \large\ \ log_bx + log_by = log_b(xy) ]
*[tex \large\ \ log_bx - log_by = log_b(x/y) ]