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You are given the following equation between a worker's nominal wage in natural log, ln( Wage), and experience, where experience (henceforth, Exp) is defined as the number of years that have elapsed since this worker finished schooling: ln(W age )=6.9+0.05Exp For example, for a fresh graduate (e.g., Exp=0 ), his/her wage (in natural logs) is ln( Wage) = 6.9+0.05∗0=6.9. His/her wage in levels is e6.9=992 dollar per month. 1. What is this worker's wage in natural logs after working for 6 years? 2. What is this worker's wage in levels? Denote your answer as W6 3. By what percentage would this worker's wage increase after working for an additional year, from 6 to 7 ? Explain your reasoning. 4 .Labor economists find experience does not increase a worker's wage monotonically. Moreover, additional experience might negatively affect the wage after a worker's experience exceeds some "threshold". Taking this view into account, they estimate a quadratic equation between experience and ln( Wage). ln( Wage )=6.9+0.05Exp−0.001Exp2 Under this equation, (a) What is the impact of one additional year of experience on wage from 5 to 6 ? (Hint: take the derivative of ln( Wage) with respect to Exp ) (b) At what level of experience does this worker's wage attain its maximum?

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Final answer:

The worker's wage in natural logs after 6 years is 7.2. The wage increase percentage from the 6th to the 7th year is approximately 1.72%. The quadratic model predicts the wage is maximized at 25 years of experience.

Step-by-step explanation:

The wage equation given uses experience as the variable to determine the natural logarithm of a worker's wage. To find this worker's wage after working for 6 years in natural logs, we substitute 6 into the equation:

ln(Wage) = 6.9 + 0.05∙6 = 7.2.

In levels, the worker's wage, denoted as W6, is:

W6 = e^7.2 = 1341.64 dollars per month (approximately).

To calculate the wage increase percentage from the 6th to the 7th year, we evaluate the natural logarithm at both points and find their exponential:

ln(Wage) at 6 years: 7.2

ln(Wage) at 7 years: 6.9 + 0.05∙7 = 7.25

Wage at 6 years: e^7.2

Wage at 7 years: e^7.25

The percentage change is then (e^7.25/e^7.2 - 1) × 100% ≈ 1.72%.

When considering a quadratic equation for wage, the derivative of ln(Wage) with respect to Exp shows the marginal effect of experience on wages. For the quadratic equation ln(Wage) = 6.9 + 0.05Exp - 0.001Exp^2, the derivative is:

d(ln(Wage))/dExp = 0.05 - 0.002∙Exp.

The impact of one additional year of experience from 5 to 6 is given by:

d(ln(Wage))/dExp|_{Exp=5} = 0.05 - 0.002∙5 = 0.04.

To find the level of experience at which this worker's wage is maximized, we set the derivative to zero and solve:

0 = 0.05 - 0.002∙Exp,

Exp = 0.05 / 0.002 = 25 years.

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