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

1 Answer
Oct 25, 2015

Answer:

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

Explanation:

#[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))#