# 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

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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.

**The correct answer is b.**

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

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

`Thus`

`0<x<1`

`0<log_a b<1`

**THus correct answer is B**