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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

Grade:12

2 Answers

Rinkoo Gupta
askIITians Faculty 80 Points
9 years ago

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

Nikhil Jena
12 Points
8 years ago
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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