For a >0,!=1, the roots of the equation ...

For `a >0,!=1,` the roots of the equation `(log)_(a x)a+(log)_x a^2+(log)_(a^2x)a^3=0` are given

`a^(-4//3)`

`a^(-3//4)`

`a^(-1//2)`

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The correct Answer is:

A, D

` log_(ax)a+log_(x)a^(2)+log_(a^(2)x)a^(3) = 0`
` or 1/(log_(a)ax)+2/(log_(a)x)+3/((log_(a)a^(2)x))= 0` `or 1/(1+log_(a)x) + 2/(log_(a) x) + 3/((2+log_(a)x )) = 0`
Let ` log_(a)x = y ," we have" 1/(y+1) + 2/y+3/(2+y) = 0`
` or 6y^(2) + 11y+4 = 0`
` or y = log_(a) x = - 1/2, - 4/3`
` rArr x = a^(-4//3), a^(-1//2)`

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If a, b and c are positive numbers in a G.P., then the roots of the quadratic equation (log_ea)x^2-(2log_eb)x.+(log_ec)=0 are

`-1 and (log_ec)/(log_ea)`

`1 and- (log_ec)/(log_ea)`

`1 and log_ac`

`-1 and log_ca`

For `a >0,!=1,` the roots of the equation `(log)_(a x)a+(log)_x a^2+(log)_(a^2x)a^3=0` are given

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If a, b and c are positive numbers in a G.P., then the roots of the quadratic equation (log_ea)x^2-(2log_eb)x.+(log_ec)=0 are

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