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A painting sold for $212 in 1975 and was sold again in 1990 for $494. Assume that the growth in the value V of the collector's item was exponential.

a) Find the value k of the exponential growth rate. Assume V₂ = 212.
k=
(Round to the nearest thousandth.)
b) Find the exponential growth function in terms of t, where t is the number of years since 1975.
v(t) =
c) Estimate the value of the painting in 2005.
$
(Round to the nearest dollar.)
d) What is the doubling time for the value of the painting to the nearest tenth of a year?
years
(Round to the nearest tenth.)
e) Find the amount of time after which the value of the painting will be $1851.
years
(Round to the nearest tenth.)

A painting sold for $212 in 1975 and was sold again in 1990 for $494. Assume that-example-1
User Frishi
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1 Answer

5 votes

Final answer:

a) The value k of the exponential growth rate is approximately 0.042. b) The exponential growth function is V(t) = 212 * e^(0.042*t). c) The estimated value of the painting in 2005 is $683. d) The doubling time is approximately 16.4 years. e) The time when the value of the painting will be $1851 is approximately 42.4 years.

Step-by-step explanation:

a) Find the value k of the exponential growth rate:

To find the exponential growth rate, we can use the formula V(t) = V₀ * e^(kt), where V(t) is the value at time t, V₀ is the initial value, k is the growth rate, and e is the base of natural logarithms.

Since V₂ = 212 and t₂ - t₁ = 1990 - 1975 = 15, we have:

212 = V₀ * e^(k * 15)

Next, we can use the fact that ln(e^x) = x, where ln is the natural logarithm, to solve for k:

k = ln(212 / V₀) / 15

Rounding to the nearest thousandth, we find k ≈ 0.042.

b) Find the exponential growth function:

The exponential growth function in terms of t can be written as: V(t) = V₀ * e^(kt), where V(t) is the value at time t, V₀ is the initial value, k is the growth rate, and e is the base of natural logarithms.

So, the exponential growth function for this problem is: V(t) = 212 * e^(0.042*t).

c) Estimate the value in 2005:

To estimate the value in 2005, we can substitute t = 2005 - 1975 = 30 into the exponential growth function:

V(30) = 212 * e^(0.042 * 30)

Rounding to the nearest dollar, the estimated value in 2005 is $683.

d) Calculate the doubling time:

The doubling time can be calculated using the formula t₂ - t₁ ≈ (ln(2) / k), where t₁ is the initial time, t₂ is the final time, and k is the growth rate.

For this problem, t₂ - t₁ = 1990 - 1975 = 15, so:

15 ≈ (ln(2) / 0.042)

Rounding to the nearest tenth, the doubling time is approximately 16.4 years.

e) Find the time when the value is $1851:

To find the time when the value is $1851, we can substitute V(t) = 1851 into the exponential growth function:

1851 = 212 * e^(0.042 * t)

Dividing both sides by 212 and taking the natural logarithm, we have:

ln(1851/212) = 0.042 * t

Dividing both sides by 0.042, we find:

t ≈ ln(1851/212) / 0.042

Rounding to the nearest tenth, the time when the value is $1851 is approximately 42.4 years.

User Winter Soldier
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