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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)` |
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Answer» Correct Answer - A::D `log_(ax)(a) +log_(x)(a^(2))+log_(a^(2)x)a^(3) = 0` `rArr (1)/(log_(a)(ax))+(2)/(log_(a)x)+(3)/(log_(a)(a^(2)x)=0` `rArr (1)/(1+log_(a)x)+(2)/(log_(a)x)+(3)/(2+log_(a)x)=0` Let `log_(a)x = t` `(1)/(1 + t) + (2)/(t) + (3)/(2 + t) = 0` `rArr t(2+t) + 2(1+t)(2+t) + 3t(1+t)=0` `rArr 2t+t^(2)+2(t^(2)+3t+2)+3t^(2)+3t=0` `rArr 6t^(2)+11t+4=0` `rArr 6t^(2)+8t+3t+4=0` `rArr 2t(3t + 4) + 1(3t + 4) = 0` `rArr (3t + 4)(2t + 1) = 0` `:. t = -(4)/(3)` or `t = -(1)/(2)` `log_(a)x = -(4)/(3)` or `log_(a)x = -(1)/(2)` `x = a^(4//3), x = a^(-1//2)` |
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