If a, b and c are positive numbers in a GP, then the roots of the quadratic equation `(log_e a)^2-(2log_e b)x+(log_ec)=0` are

Options

A. ` -1 and (log_(e)c)/(log_(e)a)`
B. ` 1 and = (log_(e)c)/(log_(e)a)`
C. ` 1 and log_(a)c`
D. ` =1 and log_(c)a`

Answer

c

Explanation

Since a,b and c are in GP
` :. B^(2) = ac`
Given equation is
` (log_(e)a)x^(2) -(2 log_(e)b)x +(log_(e)C) =0`
On putting x =1, we get
`log_(e)a -2 log_(e) b+log_(e)c=0`
` rArr 2 log_(e)b = log_(e)a+log_(e)c`
` rArr log_(e)b^(2) = log_(e)ac`
`rArr b^(2) = ac, ` which is true .
Hence , one of the root of given equation is 1
Let another root be `alpha`
` :. "Sum of roots " , = (2 log _(e)b)/(log_(e)a) = (log_(e)b^(2))/(log_(e)a)`
` rArr alpha = (log_(e)ac)/(log_(e)a) -1` lt
` = ((log_(e)a+log_(e)c))/(log_(e)a) -1`
` = (log_(e)c)/(log_(e)a)= log_(a)c`
Hence , roots are 1 and `log_(a)` C.