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

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Answer 1 · 2016-04-22T11:05:59

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

We will use the following:

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#