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

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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`

Posted by user8235304 on July 4, 2013 at 5:49 PM via web and tagged with calculus, math

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.

Posted by ishpiro on July 4, 2013 at 6:09 PM (Answer #1)