How do you express a difference of logarithm ${log}_{a} \cdot (\frac{x}{y})$ $log_a *(x/y)$ ?

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Answer 1 · 2015-06-19T19:37:22

${log}_{a} (\frac{x}{y}) = {log}_{a} x - {log}_{a} y$ $log_a(x/y)= log_ax-log_ay$

This is gotten from the result ${log}_{b} (\frac{A}{B}) = {log}_{b} A - {log}_{b} B$ $log_b(A/B)= log_bA-log_bB$

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

$\Rightarrow {log}_{b} A = m$ $=>log_bA=m" "$ and ${log}_{b} B = n$ $" "log_bB=n$

From the first statement, $\frac{A}{B} = \frac{b^{m}}{b^{n}}$ $A/B= b^m/b^n$

And by law of indices,

$\Rightarrow \frac{A}{B} = b^{m - n}$ $=> A/B= b^(m-n)$
$\Rightarrow {log}_{b} (\frac{A}{B}) = m - n$ $=> log_b(A/B)= m-n$

Recall that, $\Rightarrow {log}_{b} A = m$ $=>log_bA=m" "$ and ${log}_{b} B = n$ $" "log_bB=n$

$\Rightarrow {log}_{b} (\frac{A}{B}) = {log}_{b} A - {log}_{b} B$ $=> color(blue)(log_b(A/B)= log_bA-log_bB)$

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