Question

( 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:

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

Solution