Question 130285
let x="months until same price"


92(1.12)^(x/6)=80(1.15)^(x/4) __ dividing by 80 __ 1.15(1.12)^(x/6)=(1.15)^(x/4)


taking log __ log(1.15)+(x/6)(log(1.12))=(x/4)(log(1.15))


subtracting (x/6)(log(1.12))__ log(1.15)=(x/4)(log(1.15))-(x/6)(log(1.12))


factoring __ log(1.15)=x[(log(1.15))/4-(log(1.12))/6]


dividing by [(log(1.15))/4-(log(1.12))/6] __ (log(1.15))/[(log(1.15))/4-(log(1.12))/6]=x