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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^2a)a^3=0` are given `a^(-4/3)` (b) `a^(-3/4)` (c) `a` (d) `a^(-1/2)`

A

`a^(-4//3)`

B

`a^(-3//4)`

C

a

D

`a^(-1//2)`

Text Solution

Verified by Experts

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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