# How do you calculate #log_(1/625) 125#?

##### 1 Answer

Apr 22, 2016

#### Answer:

#### Explanation:

We will use the following:

#y = log_a(x) <=> a^y = x# (for#x>0# )#log_a(a^x) = x# #(a^b)^c = a^(bc)# #a^(-b) = 1/a^b#

Let