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# When you evaluate any `log_a b` where a > b, a `!=` 1, then a)  `log_a b` > 1...

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When you evaluate any `log_a b` where a > b, a `!=` 1, then

a)  `log_a b` > 1

b)  0 < `log_a b` < 1

c)  `log _ ab` < 0

d)  Cannot be determined

Posted by kristenmariebieber on August 18, 2013 at 3:53 AM via web and tagged with algebra2, math

College Teacher

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When `a!=1` and a > b, `log_a b` can take on any value for different values of a and b.

`log_a b` can take on all values including 0 as there is no restriction in place that does not allow b to take on the value 1.

For example:

For a = 12, b = 10, `log_12 10 ~~ 0.9266`

For a = 0.8, b = 0.6, `log_0.8 0.6 ~~ 2.2892`

For a = 8, b = 0.1, `log_ 8 0.1 ~~ -1.107`

For a = 8, b = 1, `log_8 1 = 0`

The correct option is D.

Posted by justaguide on August 18, 2013 at 4:23 AM (Answer #1)

High School Teacher

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Given

`a>b```  so  `ln(a)>ln(b)`

`1>(ln(b))/(ln(a))`     (i)

`log_a b=x`

`a^x=b`

`xln(a)=ln(b)`

`x={ln(b)}/{ln(a)}<1`       (ii)

`=> x in(-oo,1)`

`=> log_a b in(-oo,1)`

Thus option A is not possible

Options B and C are correct, so we can not decide either B or C.

Thus correct answer for this multiple choice question will be D.

Posted by aruv on August 18, 2013 at 5:17 AM (Answer #2)