# Q. `a = log_(1/2)sqrt0.125` and `b = log_3{1/(sqrt24 - sqrt 17)}` then a) `a>0 and b>0` b)`a<0 and b<0` c)`a>0 and b<0` d)`a<0 and b>0`

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### 1 Answer

**The answer is a): a > 0 and b > 0**

A logarithm is positive when the base is greater than 1 and the argument is greater than 1, and it is also positive when the base is less than 1 and the argument is less than 1.

This is because a number greater than 1 taken to a positive power remains greater than 1, and a number less than 1 taken to a positive power remains less than 1.

In logarithm *a*, base `1/2` is less than 1 and the argument `sqrt(0.125)` is less than 1. Therefore *a* is positive.

In logarithm *b*, base 3 is greater than 1 and the argument is the reciprocal of `sqrt(24) - sqrt(17)` . Square root of 24 is a bit less than 5 (square root of 25) and square root of 17 is a bit greater than 4 (square root of 16.) So the difference is less than 1 and thus reciprocal is greater than 1. Therefore *b* is positive as well.