Solve : log 6 x + logox = log 18 3-...
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Solve : log 6 x + logox = log 18 3-1

JEE/Engineering Exams
Maths
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( 2 log _{10} pi ) ( operatorname{mog}_{10}^{2} x+log _{10} x^{2}=log _{10}^{2} 3-1 ) Let ( log _{10} x=t ) ( Rightarrow quad t^{2}+2 z=left(log _{10} 3-1right)left(log _{10} 3+1right) ) ( Rightarrow x^{2}+2 t=log _{10}left(frac{3}{10}right) cdot log _{10} 30 ) ( Rightarrow quad t(t+2)=left(log _{10} 3-1right)left(log _{10} 3+1right) ) foin ebservation, we can soly that ( t=log _{10} 3-1 ) ( Rightarrow quad t+2=108,3-1+2=operatorname{cog}_{10} 3+1 ) ( therefore log _{10} x=log _{10} 3-log _{10} 10 ) ( =log _{10}left(frac{3}{10}right) ) ( operatorname{sen}^{2 alpha^{2} y} Rightarrowleft|x=frac{3}{10}right| ) Ans:
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