Problem

Compare \(P(m)=100(1.01)^m\) and \(P(y)=100[(1.01)^12]^y\) at matched elapsed times. State what \(m\) and \(y\) count, use \(m=12y\), rewrite the yearly power, and justify whether the models have the same initial value and outputs.

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What this problem is really about

Models using different time units must be compared at the same elapsed time, not at equal numerical inputs. We’ll relate the month and year counts, substitute the matched count into the monthly model, group the monthly factors into annual factors, and compare the resulting expression and initial value with the yearly model.

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They are equivalent because both start at 100. No input mapping, interval-factor comparison, or algebra establishes equivalence.

They are not equivalent because setting m = y gives different values. The models use different time units, so matched times require m = 12y.

They are equivalent at matched times m = 12y. The monthly factor 1.01 becomes yearly factor 1.01¹2, and 100(1.01)¹²y = 100[(1.01)¹2]y.

They are not equivalent because the yearly factor is 12(1.01) = 12.12. Repeated monthly growth uses the power 1.01¹2, not multiplication by 12.

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Compare \(P(m)=100(1.01)^m\) and \(P(y)=100[(1.01)^12]^y\) at matched elapsed times. State what \(m\) and \(y\) count, use \(m=12y\), rewrite the yearly power, and justify whether the models have the same initial value and outputs.

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