log5 120 + (x-3)-2. log5 (1-5^x-3)=-log5 (0.2-5^x-4) find x

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Rinkoo Gupta · Answer

log5 120 + (x-3)-2. log5 (1-5^x-3)=-log5 (0.2-5^x-4) log 5 120+log 5 (0.2-5 x-4 )-log 5 (1-5 x-3 ) 2 =3-x log5[120(0.2-5 x-4 )/(1-5 x-3 )]=3-x [120(0.2-5 x-4 )/(1-5 x-3 )]= 5 3-x 24-(24/125).5 x =5 3-x -1 25=125/5 x +(24/125).5 x Let 5 x =y, then 25=125/y +(24/125).y 24y 2 -(125*25)y+(125)(125)=0 Solving by shridharacharya method we get Y=125, 125/24 Taking y=125 5 x =125=5 3 X=3 Ans. Thanks & Regards Rinkoo Gupta AskIITians Faculty

Nikhil Jena · Answer

3 is not the correct answer. Please verify the question. Because in log5 (1-5^x-3) if we place 3 then output will come log5 0. And also the correct answer is 1.

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log5 120 + (x-3)-2. log5 (1-5^x-3)=-log5 (0.2-5^x-4) find x

log5 120 + (x-3)-2. log5 (1-5^x-3)=-log5 (0.2-5^x-4) find x

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