Dear Student,
Taking Log on both sides,u get xlog1/logi = xlog2
U have log1=0 and log2 is not equal to 0,which means x has o be zero
and that is the only integral solution
THANKS
ALL THE BEST
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Dear Student,
Sorry for the wrong reply,i went wrong somewhere...
Anyways,taking Log on both sides,u get xLog(1-i) = x Log2
multiply and divide 1-i by root 2,u get 1-i/2^0.5 = e^-ipie/4
that is x( 0.5Log2 - i22/28 ) = xLog2
that is xLog2 = -i22/28 and x=0 are the solutions
clearly,x is imaginary if the 1st eqn is true,hence,x=0 is the only integral solution
 
THANKS
ALL THE BEST
PLZZZ APPROVE MY ASNWER IF YOU LIKE IT
						

							(1-i)^x=2^x1-iComplex number is of the from x+iyWhere x=1             Y=-1I 1-i l=√x^2+y^2l 1-i l=√2 Hence  l 1-i l^x=2^x  Putting value     (√2)^x=2^x     √2^x=(√2×√2)^x    √2^x=(√2)^2x  Comparing powers           x=2x           x-2x=0            -x=0              X=0
						

							
Taking log on both sides, we get
 
x log(1-i) = x log2
 
Now multiplying and dividing the (1-i) part with (1+i) to get:
x {log 2 - log (1+i) } =  x log 2
This is only possible if x =0. Hence only one integral solution exist.
 
Correct me if am wrong.
Spandan Mallick, JEE Aspirant