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

### 1 Answer | Add Yours

`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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