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

1 Answer
Jun 19, 2015

Answer:

#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)#