1.

`log_100 a-log_a 0.0001 , a >1`A. 2B. 3C. 4D. none of these

Answer» Correct Answer - C
We have,
`2log_(100)a-log_(a)0.0001`
`=2log_(100)a-log_(a)(100)^(-2)`
`=2log_(100)a+2log_(a)100`
`=2(log_(100)a+log_(a)100)`
`=(log_(100)a+(1)/(log_(100)a))`
`ge2xx2=4" "[becausex+(1)/(x)ge2" for all "xgt0]`


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