Question

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

Answer 1

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

Explanation:

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)`