(hanging into log_(2)(x^(2))+log_(4)y=7 base 10. (1) Then{:[=(2log x)/(log 2)+(log y)/(log 4)=7quad:'[[log_(10)x=log x],[log x^(n)=n log x]]],[=(2log x)/(0.30)+(log y)/(0.60)=7quad:'[[log 2=0.30],[log 4=0.60]]],[=6.66 log x+1.66 log y=7]:}Criven II equation (log_(10)(8x))/(log_(10)4)+(log_(4)(8x)+log_(20)(2y^(2)2))/(log_(10)2)=8 {:=>(log(8x))/(log 4)+(log 2y^(2))/(log 2)=8quad:'log_(b)a=log_(10)9] {:log_(10)k] (log 8x)/(0.60)+(2log(2y))/(0.30)=8 =1.66 lo ... See the full answer