find the integral solutions of the equation (1-i)^x=(2)^x

Question

Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad · Answer

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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Sukhendra Reddy Rompally B.Tech Mining Machinery Engg, ISM Dhanbad · Answer

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

mycroft holmes · Answer

Vishva vivek singh · Answer

(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

Spandan Mallick · Answer

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

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find the integral solutions of the equation (1-i)^x=(2)^x

find the integral solutions of the equation (1-i)^x=(2)^x

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