How do you express a difference of logarithm $log_a *(x/y)$ ?

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Answer 1 · 2015-06-19T19:37:22

$log_a(x/y)= log_ax-log_ay$

This is gotten from the result $log_b(A/B)= log_bA-log_bB$

If you're curious as to how this is possible then continue reading

Suppose, $b^m= A" "$ and $" "b^n=B$

$=>log_bA=m" "$ and $" "log_bB=n$

From the first statement, $A/B= b^m/b^n$

And by law of indices,

$=> A/B= b^(m-n)$
$=> log_b(A/B)= m-n$

Recall that, $=>log_bA=m" "$ and $" "log_bB=n$

$=> color(blue)(log_b(A/B)= log_bA-log_bB)$

How do you express a difference of logarithm log_a *(x/y)log_a *(x/y)?