Problem

Compare savings $R(t)=120\cdot~1.03^{t}$ and $S(t)=90\cdot~1.06^{t}$ by giving their starts and yearly growth rates, solving for the crossover time, and identifying the greater balance before and after it.

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starts = R=$120, S=$90; growth = R 3%, S 6%; crossover = none; comparison = S>R initially because growth rate is higher

starts = R(0)=$120, S(0)=$90; growth = R 3%/year, S 6%/year; crossover = (1.06/1.03)ᵗ=4/3 ⇒ t≈10.020 years; comparison = R>S before crossover, equal there, S>R after

starts = R=$120, S=$90; growth = R 3%, S 6%; crossover = t=10 exactly from 30/3; comparison = R before, S after

starts = R=$120, S=$90; growth = R 3%, S 6%; crossover = none; comparison = R remains larger forever because it starts larger

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Compare savings \(R(t)=120\cdot~1.03^{t}\) and \(S(t)=90\cdot~1.06^{t}\) by giving their starts and yearly growth rates, solving for the crossover time, and identifying the greater balance before and after it.

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