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If log(0.3)(x-1) lt log(0.09)(x-1), then...

If `log_(0.3)(x-1) lt log_(0.09)(x-1)`, then x lies in the interval :

Text Solution

Verified by Experts

We have,
` log_(0.3) (x - 1) lt log_(0.09) (x - 1)= log_((0.3)^(2)) (x -1)`
`rArr log_(0.3) (x-1) lt (1)/(2) log_(0.3) (x -1)`
` rArr 2 log_(0.3) (x -1) lt log_(0.3) (x-1)` …(i)
`rArr 2 log_(0.3)(x-1) - log_(0.3) (x-1) lt 0 `
`rArr log_(0.3) (x-1) lt 0 `
`rArr (x -1) gt(0.3)^(0) = 1
`rArr x gt 1 + 1 = 2 `
therefore x in (2, oo)`
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