Let us investigate the complex solutions. We know by the logarithm definition
4^z = -1/64 = -4^{-3}4z=−164=−4−3
Supposing now z = x + i yz=x+iy we have
4^x 4^{iy} = -4^{-3}4x4iy=−4−3 so we have
4^x = 4^{-3}->x = -34x=4−3→x=−3 and
4^{iy} = -14iy=−1
We know
4 = e^{log_e 4}4=eloge4 so
4^{iy} = e^{i y log_e 4} = cos(y log_e 4)+i sin(y log_e 4) = -14iy=eiyloge4=cos(yloge4)+isin(yloge4)=−1. This condition is attained for
y log_e4 = pi pm 2kpiyloge4=π±2kπ with k= 0,1,2,3,4,cdotsk=0,1,2,3,4,⋯
so
y = (pi pm 2kpi )/log_e 4y=π±2kπloge4
Finally
log_4(-1/64) = -3+i (pi pm 2kpi )/log_e 4log4(−164)=−3+iπ±2kπloge4
