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# find the integral solutions of the equation (1-i)^x=(2)^x Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad
93 Points
10 years ago

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

PLZZZ APPROVE MY ANSWER IF YOU LIKE IT Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad
93 Points
10 years ago

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

10 years ago 3 years ago
(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
2 years ago
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