Lets Find Interval

Algebra Level 2

If log 0.3 ( x 1 ) < log 0.09 ( x 1 ) \displaystyle \log _{ 0.3 }{ \left( x-1 \right) } <\log _{ 0.09 }{ \left( x-1 \right) } , then x x lies in the interval __________ . \text{\_\_\_\_\_\_\_\_\_\_}.

2 < x < 2<x<\infty 2 < x < 1 -2<x<-1 < x < 2 -\infty <x<2 1 < x < 2 1<x<2

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1 solution

Soumo Mukherjee
Mar 18, 2015

Transforming the given equation

log 0.3 ( x 1 ) < log ( 0.3 ) 2 ( x 1 ) = 1 2 log 0.3 ( x 1 ) 1 2 log 0.3 ( x 1 ) < 0 o r log 0.3 ( x 1 ) < 0 \log _{ 0.3 }{ \left( x-1 \right) } <\log _{ { \left( 0.3 \right) }^{ 2 } }{ \left( x-1 \right) } =\cfrac { 1 }{ 2 } \log _{ 0.3 }{ \left( x-1 \right) } \\ \therefore \quad \cfrac { 1 }{ 2 } \log _{ 0.3 }{ \left( x-1 \right) } <0\\ or\quad \log _{ 0.3 }{ \left( x-1 \right) } <0

( x 1 ) > 1 \left( x-1 \right) >1 or x > 2 x>2\quad . As the base is less than 1 1 the inequality is reversed. Hence the answer.

Yes, I too did the same! Expressed base 0.09 in terms of base 0.3.

Venkata Karthik Bandaru - 6 years, 2 months ago

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