Problem

Are these \(\text{two}\) rearrangements equivalent? \(r=(A/P)^{1/t}-1\) and \(1+r=(A/P)^{1/t}\)

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What this problem is really about

Test equivalence by transforming one formula into the other with a reversible operation on both sides. The subtraction occurs outside the power, so adding the same constant to both sides cancels it without changing the exponent. If the second display appears exactly and the operation can be reversed on the shared domain, the rearrangements describe the same relationship.

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No, they are not equivalent because the -1 is part of the exponent, so adding 1 would change the power.

No, they are not equivalent because adding 1 to the left side should make r+1=1+(A/P)¹/t-1, not 1+r=(A/P)¹/t.

No, they are not equivalent because moving the 1 across the equals sign should make r=(A/P)¹/t+1 instead.

Yes, they are equivalent because adding 1 to both sides of r=(A/P)¹/t-1 gives 1+r=(A/P)¹/t.

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Are these \(\text{two}\) rearrangements equivalent? \(r=(A/P)^{1/t}-1\) and \(1+r=(A/P)^{1/t}\)

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