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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) log5120+log5(0.2-5x-4)-log5(1-5x-3)2=3-x log5[120(0.2-5x-4)/(1-5x-3)]=3-x [120(0.2-5x-4)/(1-5x-3)]= 53-x 24-(24/125).5x=53-x -1 25=125/5x+(24/125).5x Let 5x=y, then 25=125/y +(24/125).y 24y2-(125*25)y+(125)(125)=0 Solving by shridharacharya method we get Y=125, 125/24 Taking y=125 5x=125=53 X=3 Ans.Thanks & RegardsRinkoo GuptaAskIITians Faculty
log5 120 + (x-3)-2. log5 (1-5^x-3)=-log5 (0.2-5^x-4)
log5120+log5(0.2-5x-4)-log5(1-5x-3)2=3-x
log5[120(0.2-5x-4)/(1-5x-3)]=3-x
[120(0.2-5x-4)/(1-5x-3)]= 53-x
24-(24/125).5x=53-x -1
25=125/5x+(24/125).5x
Let 5x=y, then
25=125/y +(24/125).y
24y2-(125*25)y+(125)(125)=0
Solving by shridharacharya method we get
Y=125, 125/24
Taking y=125
5x=125=53
X=3 Ans.
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
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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