Question

161. logx (9x”). logź x = 4. ( 2)

JEE/Engineering Exams

Maths

Solution

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( log _{x}left(9 x^{2}right) cdot log _{3}^{2} x ) As ( left.log _{a} b=frac{log b}{log a}right) ) ( Rightarrow frac{log 9 x^{2}}{log x} timesleft(frac{log x}{log 3}right)^{2}=4 ) (ss log ab ( =text { wogat log b }) ) ( Rightarrowleft(frac{log 9+log x^{2}}{(log 3)^{2}} log x=4right. ) ( log _{(log x)^{2}+3left(log _{3} 2right) log x-4(log 3)^{2}=0} ) ( log x=-log _{frac{1}{2}} frac{10 y^{2} 3+8 log ^{2} 3}{2} ) ( therefore x=frac{1}{9}, 3 )