Question

( log _{a} b=2 )
o) ( b=a^{2}-c i )
( log _{b} c=2 )
( Rightarrow c=b^{2}-c i )
( log _{3} c=3+log _{3} a )
( Rightarrow c=3^{3}-3^{log _{3} a} )
( =a cdot 27-operatorname{cin} )
( a cdot 27=b^{2}- ) from ( quad ) is
( Rightarrow a=3 quad t=sqrt{2} b=9, quad c=81 )
( a+b+c=93 )

# WD) TV (IU (D) TU 7. If log b = 2; log c = 2 and log c = 3 + log,a then (a + b + c) equals (A) 90 (B)93 (C) 102 (D) 243

Solution