How do you expand #log (1/ABC) #?

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Answer 1 · 2015-10-25T13:17:48

#log(1/ABC)=-log(A)+log(B)+log(C)#

#[1]" "log(1/ABC)#

Property: #log_b(mn)=log_b(m)+log_b(n)#

#[2]" "=log(1/A)+log(B)+log(C)#

Property: #log_b(m/n)=log_b(m)-log_b(n)#

#[3]" "=log(1)-log(A)+log(B)+log(C)#

Property: #log_b(1)=0#

#[4]" "=color(blue)(-log(A)+log(B)+log(C))#