Question 175313
The subscript next to log specifies the "base".
.
log_3(21)- log_3(7)
applying "log rules":
log_3(21)- log_3(7) = log_3(21/7) = log_3(3) = 1
Note: since 3^1 = 3
.
log(4+x)-log(x-5)=log2 
log((4+x)/(x-5))=log2 
(4+x)/(x-5) = 2 
(4+x) = 2(x-5)
4+x = 2x-10
4 = x-10
14 = x
.
To review log rules see:
http://www.purplemath.com/modules/logrules.htm