Iflog0.3(x-1)​<log0.09(x-1) then x lies in the interval

I can understand up to ---​2log0.3(x-1)<log0.3(x-1)
(x-1)2>(x-1) --- next step sign changed from < to >. Can you please explain sir

Dear Student,
Please find below the solution to the asked query:

Note that logax1>logax2 givesx1>x2if a>1x1<x2if 0<a<1i.e. sign of inequality reverses if base lies between 0 and 1Also x1>0 and x2>0Nowlog0.3 x-1<log0.09x-1Quantity inside log is always positive, hencex-1>0x1,...ilog0.3 x-1<log0.09x-1log0.3 x-1<log0.32x-1log0.3 x-1<12log0.3x-1 As loganb=1n logab2log0.3 x-1<log0.3x-1 log0.3 x-12<log0.3x-1 As logabm=m logabNow using x1>x2if a>1x1<x2if 0<a<1, we getx-12>x-1x-12-x-1>0x-1x-1-1>0x-1x-2>0Now using method of interval-+++++1----2+++++Hence x-,12,...iiiii givesx2,

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