91Ó°ÊÓ

First, let's analyze the given inequality. Notice that for \(l = 1\), the inequality reduces to \((1-d) < 1-dt\) which is true since \(d > 0\). Now, for \(l > 1\), we consider that the term \((1-d)^l\) is a monotonically decreasing function due to \(1-d\) being between 0 and 1. This means that the term \((1-d)^l\) will be smaller than \((1-d)\), and since \(t < 1\), it means that \(dt < d\). Thus, we can conclude that \((1-d)^l < 1-dt\) for \(0 < t < 1\).

Now, let's analyze the given equation for this case. We want to show that for \(t = 1\), the two expressions are equal. Let's examine the left-hand side: \((1-d)^f = (1-d)^1 = 1-d\) Now, let's look at the right-hand side: \(1 - dt = 1-d\) Since both expressions are equal, we have proven that \((1-d)^f = 1-dt\) for \(t = 1\).

Finally, let's analyze the inequality for this case. We want to show that if \(t > 1\), then \((1-d)^\prime > 1-dt\). As mentioned earlier, the term \((1-d)\) is between 0 and 1 which makes \((1-d)^\prime\) a monotonically increasing function as the exponent \(\prime\) increases. Consequently, for any value of \(t > 1\), the term \((1-d)^\prime\) will be greater than \((1-d)\). On the other hand, the term \(1-dt\) will decrease as \(t\) increases for any value of \(t > 1\). Thus, we can deduce that \((1-d)^\prime > 1-dt\) for \(t > 1\). In conclusion, we have successfully proved the given inequalities and equalities for all the necessary cases, showing the relative magnitudes of present values at simple and compound discount over various periods of time.