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

1 Answer
Apr 22, 2016

Answer:

#log_(1/625)(125)= -3/4#

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 #x = log_(1/625)(125)#

#=> (1/625)^x = 125#

#=> (5^(-4))^x = 5^3#

#=> 5^(-4x) = 5^3#

#=> log_5(5^(-4x)) = log_5(5^3)#

#=> -4x = 3#

#:. x = -3/4#