How do you evaluate #log_.25 (-4)#?

1 Answer
George C.
Aug 12, 2016

The principal value of #log_(0.25)(-4)# is #-1-pi/(ln(4))i#

Explanation:

Use the change of base formula to find:

#log_(0.25)(-4) = ln(-4)/ln(0.25) = ln(-4)/(ln(1/4)) = ln(-4)/(-ln(4)) = (ln(4)+pi i)/(-ln(4))#

#= -1-pi/(ln(4)) i#

Note this is the principal value of the logarithm.

There are other solutions of #0.25^x = -4#, namely:

#x = -1-(pi(1+2k))/(ln(4))i#

for any integer #k#.

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