the purchasing power (in dollars) of an annual amount of A dollars after t years of 5% inflation decays according to

P= Ae^-0.05t

a) how long will it be before a pension of $60,000 per year has a purchasing power of $30,000

b) how much pension A would be needed so that the purchasing power P is $50,000 aftter 15 years?

To find the answers to these questions, we need to use the given formula for the purchasing power:

P = Ae^(-0.05t)

a) To determine how long it will take for a pension of $60,000 per year to have a purchasing power of $30,000, we need to solve for t in the equation:

$30,000 = $60,000 * e^(-0.05t)

To isolate t, divide both sides of the equation by $60,000:

$30,000 / $60,000 = e^(-0.05t)

0.5 = e^(-0.05t)

To eliminate the exponential, we can take the natural logarithm (ln) of both sides:

ln(0.5) = ln(e^(-0.05t))

Now, we can use the logarithmic property that ln(e^x) = x:

ln(0.5) = -0.05t

Solve for t by dividing both sides by -0.05:

t = ln(0.5) / (-0.05)

Using a calculator or math software, evaluate this expression:

t ≈ 13.86 years

Therefore, it will take approximately 13.86 years for a pension of $60,000 per year to have a purchasing power of $30,000.

b) To determine the pension amount A needed for a purchasing power of $50,000 after 15 years, we can rearrange the formula:

P = Ae^(-0.05t)

$50,000 = A * e^(-0.05 * 15)

Divide both sides by e^(-0.05 * 15):

$50,000 / e^(-0.05 * 15) = A

Using a calculator or math software, evaluate this expression to find A.

A ≈ $114,536.79

Therefore, a pension amount of approximately $114,536.79 would be needed to maintain a purchasing power of $50,000 after 15 years.

To solve the given problem, we need to use the provided equation P = Ae^(-0.05t), where P represents the purchasing power in dollars, A represents the initial amount in dollars, and t represents the number of years.

a) To find how long it will take for a pension of $60,000 per year to have a purchasing power of $30,000, we need to set P equal to $30,000 and solve for t.

P = 30,000
A = 60,000

30,000 = 60,000e^(-0.05t)

Divide both sides of the equation by 60,000:

0.5 = e^(-0.05t)

Now, take the natural logarithm of both sides to get rid of the exponential function:

ln(0.5) = ln(e^(-0.05t))

Since ln(e^(-0.05t)) simplifies to -0.05t, we can write the equation as:

ln(0.5) = -0.05t

Using a calculator, you will find that ln(0.5) is approximately -0.6931.

-0.6931 = -0.05t

Now, divide both sides of the equation by -0.05:

-0.6931 / -0.05 = t

t ≈ 13.862

Therefore, it will take approximately 13.862 years for the pension of $60,000 per year to have a purchasing power of $30,000.

b) To determine the initial pension amount A required to have a purchasing power of $50,000 after 15 years, we set P equal to $50,000 and solve for A.

P = 50,000
t = 15

50,000 = Ae^(-0.05 * 15)

Now, divide both sides of the equation by e^(-0.05 * 15):

50,000 / e^(-0.05 * 15) = A

Using a calculator, you will find that e^(-0.05 * 15) is approximately 0.4285.

50,000 / 0.4285 = A

A ≈ 116,662.156

Therefore, a pension amount of approximately $116,662.16 is required to have a purchasing power of $50,000 after 15 years.

a) solve for t

30 = 60e^(-.05t)

b) solve for a
50000 = a e^(-.05(15))