X. An automobile with seven years of use has a commercial value of $34057.7, but five years ago its value was $72250. If the value of such automobile varies exponentially with the time. (36-41)

36.- What was the value of the automobile when it was new?

37.- What will be the value of the automobile after 10 years of use?

38.- How many years of use will the automobile have with a value of $10277?

39.- After how long of use does the value of the automobile reduce to the half?

40.- After how many years of use the value of the automobile reduces to the fourth part of its initial value?

41.- After how many years of use does the value of the automobile reduce to the third part of its initial velocity?

I do not recommend using Roman numerals for calculus

To answer these questions, we need to use the formula for exponential decay:

V = V0 * e^(kt)

Where:
V is the value of the automobile after t years of use
V0 is the initial value of the automobile
e is the base of the natural logarithm (~2.71828)
k is the decay constant

Now, let's solve each question step by step.

36. To find the initial value of the automobile when it was new (V0), we can set up the equation using the given information. We know that after 7 years, the value is $34057.7. Plugging in these values into our formula:

34057.7 = V0 * e^(7k)

Since we don't know the value of k, we cannot solve this equation directly. However, we have another piece of information: five years ago (t = 5), the value was $72250. Using this information:

72250 = V0 * e^(5k)

Now we have a system of two equations with two unknowns (V0 and k). We can solve this system using simultaneous equations or substitution. However, for simplicity, we will use substitution.

Rearranging the second equation to solve for V0:

V0 = 72250 / e^(5k)

Substituting this expression for V0 in the first equation:

34057.7 = (72250 / e^(5k)) * e^(7k)

Now we can simplify and solve this equation. Divide both sides by 72250 and cancel out e^(5k) terms:

34057.7 / 72250 = e^(7k - 5k)

0.4713 = e^(2k)

Taking the natural logarithm of both sides to isolate the exponent:

ln(0.4713) = 2k

k = ln(0.4713) / 2 = -0.1851 (approximately)

Now we can substitute the value of k back into the second equation to solve for V0:

72250 = V0 * e^(5 * -0.1851)

72250 = V0 * e^(-0.9255)

Divide both sides by e^(-0.9255):

V0 = 72250 / e^(-0.9255) = 72250 / 0.3968 = $182020.16 (approximately)

Therefore, the value of the automobile when it was new is approximately $182020.16.

37. To find the value of the automobile after 10 years of use, we can use the same formula. Plugging in the values:

V = V0 * e^(kt)
V = 182020.16 * e^(-0.1851 * 10)
V ≈ 182020.16 * e^(-1.851)
V ≈ 182020.16 * 0.1564
V ≈ $28484.81 (approximately)

Therefore, the value of the automobile after 10 years of use is approximately $28484.81.

38. To find the number of years of use when the automobile has a value of $10277, we set up the equation:

10277 = 182020.16 * e^(-0.1851t)

Divide both sides by 182020.16 and isolate the exponent:

0.0564 = e^(-0.1851t)

Taking the natural logarithm of both sides:

ln(0.0564) = -0.1851t

Solve for t:

t = ln(0.0564) / -0.1851 ≈ 16.92 years

Therefore, the automobile will have a value of $10277 after approximately 16.92 years of use.

39. To find the number of years of use when the value of the automobile reduces to half, we set up the equation:

0.5 = e^(-0.1851t)

Taking the natural logarithm of both sides:

ln(0.5) = -0.1851t

Solve for t:

t = ln(0.5) / -0.1851 ≈ 3.756 years

Therefore, the value of the automobile reduces to half after approximately 3.756 years of use.

40. To find the number of years of use when the value of the automobile reduces to the fourth part of its initial value, we set up the equation:

0.25 = e^(-0.1851t)

Taking the natural logarithm of both sides:

ln(0.25) = -0.1851t

Solve for t:

t = ln(0.25) / -0.1851 ≈ 7.512 years

Therefore, the value of the automobile reduces to the fourth part of its initial value after approximately 7.512 years of use.

41. To find the number of years of use when the value of the automobile reduces to the third part of its initial value, we set up the equation:

1/3 = e^(-0.1851t)

Taking the natural logarithm of both sides:

ln(1/3) = -0.1851t

Solve for t:

t = ln(1/3) / -0.1851 ≈ 5.522 years

Therefore, the value of the automobile reduces to the third part of its initial value after approximately 5.522 years of use.