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(0.2-5^x-4)-log₅(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

24y²-(125*25)y+(125)(125)=0

Solving by shridharacharya method we get

Y=125, 125/24

Taking y=125

5^x=125=5³

X=3 Ans.

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Rinkoo Gupta

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