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# When you evaluate any log (base a) b where a > b and b> 1 then: a) log (base a)...

kristenmarieb... | Student, Grade 10 | Valedictorian

Posted August 27, 2013 at 5:49 PM via web

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When you evaluate any log (base a) b where a > b and b> 1 then:

a) log (base a) b > 1

b) 0 < log (base a) b < 1

c) log (base a) b < 0

d) Cannot be determined

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 27, 2013 at 5:59 PM (Answer #1)

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The value of `log_a b` has to be determined given that a > b and b > 1.

Expressing `log_a b` in terms of logarithms with the same base:

`log_a b = (log b)/(log a)`

As a > b > 1, log a > log b, and log a and log b are positive.

The value of `log_a b` lies between 0 and 1.

aruv | High School Teacher | Valedictorian

Posted August 27, 2013 at 7:20 PM (Answer #2)

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`x=log_a b`

`a^x=b`

`ln(a^x)=ln(b)`

`xln(a)=ln(b)`

`x=(ln(b))/(ln(a))`

since b>1 and a>b

therefor a>b>1

ln is an increasing fuction in `(1,oo)`

`ln(a)>ln(b)`

`1>(ln(b))/(ln(a))`

also

`(ln(b))/(ln(a))>0`

`Thus`

`0<x<1`

`0<log_a b<1`