Question #90403

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Answer 1 · 2017-12-04T05:58:54

I am unsure whether the logarithm is base 10 or base e.

If the logarithm is base 10 then, convert to base e as follows:

#log(-i) = ln(-i)/ln(10)#

Substitute #e^((3pi)/2i)# for #-i#:

#log(-i) = ln(e^((3pi)/2i))/ln(10)#

Because #ln(x)# and #e^x# are inverses, they cancel:

#log(-i) = (3pii)/(2ln(10))#

If the logarithm is base e then, substitute #e^((3pi)/2i)# for #-i#:

#log(-i) = ln(e^((3pi)/2i))#

Because #ln(x)# and #e^x# are inverses, they cancel:

#log(-i) = (3pii)/2#