Question
( mid 5 cdot 2 cdot 2+15 cdot frac{2^{2}}{2^{2}}=135 )
Let ( 2^{2}=t )
2) ( 30 t+frac{60}{t}=135 )
( Rightarrow quad 5 t+frac{12}{2}=27 )
( Rightarrow 6 t^{2}-27 z+12=0 )
( Rightarrow quad 6 t^{2}-24 z-3 z+12=0 )
( Rightarrow(6 z-3)(z-4)=0 )
( therefore t=frac{1}{2} sin 4 )
( begin{aligned} therefore 2^{x}=2^{-1} & text {on } 2^{x}=2^{2} therefore x=-1 & Rightarrow x=2 end{aligned} )
( therefore x=-1,2 )

6. 15.2X+1 +15.2-x+2 = 135
Solution
