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

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted August 19, 2013 at 4:12 AM via web

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When you evaluate any `log_(a)b`  where b > a, and a > 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

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aruv | High School Teacher | Valedictorian

Posted August 19, 2013 at 5:10 AM (Answer #1)

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`log_a b=x`

`a^x=b`

`=>`

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

`xln(a)=ln(b)`

`x=(ln(b))/(ln(a))`        (i)

since b >a  and a>1

therefore

`(ln(b))/(ln(a))>1`    (ii)

Thus from (i) and (ii)

`x>1`

`log_a b > 1`

Thus correct answer is a .

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