How do you evaluate #log_4 (-1/64)#?

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Answer 1 · 2016-08-09T10:58:57

#log_4(-1/64) = -3+i (pi pm 2kpi )/log_e 4#

for #k = 0,1,2,cdots#

Let us investigate the complex solutions. We know by the logarithm definition

#4^z = -1/64 = -4^{-3}#

Supposing now #z = x + i y# we have

#4^x 4^{iy} = -4^{-3}# so we have

#4^x = 4^{-3}->x = -3# and
#4^{iy} = -1#

We know

#4 = e^{log_e 4}# so

#4^{iy} = e^{i y log_e 4} = cos(y log_e 4)+i sin(y log_e 4) = -1#. This condition is attained for

#y log_e4 = pi pm 2kpi# with # k= 0,1,2,3,4,cdots#

so

#y = (pi pm 2kpi )/log_e 4#

Finally

#log_4(-1/64) = -3+i (pi pm 2kpi )/log_e 4#