Question
An antique table increases in value according to the function v(x)equals750(1.06) Superscript x dollars, where x is the number of years after 1980.
a. How much was the table worth in 1980?
b. If the pattern indicated by the function remains valid, what was the value of the table in 2005?
c. Use a table or a graph to estimate the year when this table will reach double its 1980 value.
Question content area bottom
Part 1
a. The table was worth $
enter your response here in 1980.
(Round to the nearest cent as needed.)
a. How much was the table worth in 1980?
b. If the pattern indicated by the function remains valid, what was the value of the table in 2005?
c. Use a table or a graph to estimate the year when this table will reach double its 1980 value.
Question content area bottom
Part 1
a. The table was worth $
enter your response here in 1980.
(Round to the nearest cent as needed.)
Answers
GPT-4o mini
To find the value of the table in 1980, we need to calculate \( v(0) \) since \( x = 0 \) corresponds to the year 1980.
The function for the value of the table is given by:
\[ v(x) = 750 \cdot (1.06)^{x} \]
### Part 1: Value in 1980
Substituting \( x = 0 \):
\[
v(0) = 750 \cdot (1.06)^{0} = 750 \cdot 1 = 750
\]
Thus, the value of the table in 1980 was **$750.00**.
### Part 2: Value in 2005
To find the value of the table in 2005, we need to calculate \( v(25) \) since 2005 is 25 years after 1980 (2005 - 1980 = 25).
\[
v(25) = 750 \cdot (1.06)^{25}
\]
Calculating \( (1.06)^{25} \):
\[
(1.06)^{25} \approx 4.2908
\]
Now we can calculate \( v(25) \):
\[
v(25) = 750 \cdot 4.2908 \approx 3218.10
\]
Thus, the value of the table in 2005 was approximately **$3218.10**.
### Part 3: Year when the table will reach double its 1980 value
Double the value of the table in 1980 is:
\[
2 \times 750 = 1500
\]
To find the year when the table's value reaches $1500, we set up the equation:
\[
750 \cdot (1.06)^{x} = 1500
\]
Dividing both sides by 750:
\[
(1.06)^{x} = 2
\]
Now, we can take the logarithm of both sides:
\[
\log((1.06)^{x}) = \log(2)
\]
Using the power rule of logarithms:
\[
x \cdot \log(1.06) = \log(2)
\]
Now, solve for \( x \):
\[
x = \frac{\log(2)}{\log(1.06)}
\]
Calculating this using a calculator (approximately):
\[
\log(2) \approx 0.3010
\]
\[
\log(1.06) \approx 0.0253
\]
Thus,
\[
x \approx \frac{0.3010}{0.0253} \approx 11.88
\]
Since \( x \) is the number of years after 1980, we add this to 1980:
\[
1980 + 11.88 \approx 1991.88
\]
So, the table is expected to reach double its 1980 value sometime in the **year 1992** (rounding down since we are estimating the year).
### Summary Results:
- a. The table was worth **$750.00** in 1980.
- b. The value of the table in 2005 was approximately **$3218.10**.
- c. The table will reach double its 1980 value around the year **1992**.
The function for the value of the table is given by:
\[ v(x) = 750 \cdot (1.06)^{x} \]
### Part 1: Value in 1980
Substituting \( x = 0 \):
\[
v(0) = 750 \cdot (1.06)^{0} = 750 \cdot 1 = 750
\]
Thus, the value of the table in 1980 was **$750.00**.
### Part 2: Value in 2005
To find the value of the table in 2005, we need to calculate \( v(25) \) since 2005 is 25 years after 1980 (2005 - 1980 = 25).
\[
v(25) = 750 \cdot (1.06)^{25}
\]
Calculating \( (1.06)^{25} \):
\[
(1.06)^{25} \approx 4.2908
\]
Now we can calculate \( v(25) \):
\[
v(25) = 750 \cdot 4.2908 \approx 3218.10
\]
Thus, the value of the table in 2005 was approximately **$3218.10**.
### Part 3: Year when the table will reach double its 1980 value
Double the value of the table in 1980 is:
\[
2 \times 750 = 1500
\]
To find the year when the table's value reaches $1500, we set up the equation:
\[
750 \cdot (1.06)^{x} = 1500
\]
Dividing both sides by 750:
\[
(1.06)^{x} = 2
\]
Now, we can take the logarithm of both sides:
\[
\log((1.06)^{x}) = \log(2)
\]
Using the power rule of logarithms:
\[
x \cdot \log(1.06) = \log(2)
\]
Now, solve for \( x \):
\[
x = \frac{\log(2)}{\log(1.06)}
\]
Calculating this using a calculator (approximately):
\[
\log(2) \approx 0.3010
\]
\[
\log(1.06) \approx 0.0253
\]
Thus,
\[
x \approx \frac{0.3010}{0.0253} \approx 11.88
\]
Since \( x \) is the number of years after 1980, we add this to 1980:
\[
1980 + 11.88 \approx 1991.88
\]
So, the table is expected to reach double its 1980 value sometime in the **year 1992** (rounding down since we are estimating the year).
### Summary Results:
- a. The table was worth **$750.00** in 1980.
- b. The value of the table in 2005 was approximately **$3218.10**.
- c. The table will reach double its 1980 value around the year **1992**.